exec_outcome stringclasses 1 value | code_uid stringlengths 32 32 | file_name stringclasses 111 values | prob_desc_created_at stringlengths 10 10 | prob_desc_description stringlengths 63 3.8k | prob_desc_memory_limit stringclasses 18 values | source_code stringlengths 117 65.5k | lang_cluster stringclasses 1 value | prob_desc_sample_inputs stringlengths 2 802 | prob_desc_time_limit stringclasses 27 values | prob_desc_sample_outputs stringlengths 2 796 | prob_desc_notes stringlengths 4 3k ⌀ | lang stringclasses 5 values | prob_desc_input_from stringclasses 3 values | tags listlengths 0 11 | src_uid stringlengths 32 32 | prob_desc_input_spec stringlengths 28 2.37k ⌀ | difficulty int64 -1 3.5k ⌀ | prob_desc_output_spec stringlengths 17 1.47k ⌀ | prob_desc_output_to stringclasses 3 values | hidden_unit_tests stringclasses 1 value |
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
PASSED | 8152b27658b5a55020bed2b893f28814 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class SolutionBBB {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner in = new Scanner(System.in);
int t = in.nextInt();
while (t-- > 0) {
int n = in.nextInt();
long[] arr = new long[n];
HashMap<Long,Long> map = new HashMap<Long,Long>();
for (int i = 0; i < n; i++) {
arr[i] = in.nextLong();
map.put(arr[i], (long)i+1);
}
Arrays.sort(arr);
long count = 0;
for (int i = 0; i < n - 1; i++) {
long x = map.get(arr[i]);
for (int j = i + 1; j < n; j++) {
long y = map.get(arr[j]);
if (x+y == arr[i] * arr[j])
count++;
if (arr[i] * arr[j] > 2 * n)
break;
}
}
System.out.print(count);
System.out.println();
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | a6048084cba4b546f2132f87fac68b65 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Solution{
public static void main(String[] args) throws IOException{
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
for(int i=0;i<t;i++) {
int n = Integer.parseInt(br.readLine());
int[] arr = new int[n+1];
String[] sarr = br.readLine().split(" ");
HashMap<Integer,Integer> hm=new HashMap<>();
for (int j = 1; j <= n; j++) {
arr[j] = Integer.parseInt(sarr[j - 1]);
hm.put(arr[j],j);
}
int count=0;
for(int j=2;j<=n;j++){
for(int k=1;k*arr[j]<=2*n;k++){
if(hm.containsKey(k)) {
if ((hm.get(k) == arr[j] * k - j) && hm.get(k) < j) {
count++;
}
}
}
}
/* LinkedHashMap<Integer,Integer> lhm=new LinkedHashMap<>();
lhm.put(1,arr[1]);
for(int j=2;j<=n;j++){
for(int k=1;k<j;k++){
if(lhm.get(k)==)
}
}*/
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | e9217c2897c24b906a33eef32c6874e0 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.Scanner;
public class Demo1 {
public static void main(String args[]) throws Exception{
Scanner sc=new Scanner(System.in);
int q=sc.nextInt();
for (int iq=1;iq<=q;iq++){
int n=sc.nextInt();
long ar[]= new long[n+1];
for (int i=0;i<n;i++)
ar[i+1]=sc.nextInt();
long ans=0;
for (int i=1;i<n+1;i++){
long val=i+ar[i]-(2*i)%ar[i];
while(val<=n){
if(ar[i]*ar[(int)val]==i+val)
ans++;
val+=ar[i];
}
}
System.out.println(ans);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 6ad1f8e4cbb278dd6fa6ab1ba578a204 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
public class Main {
static class Reader {
static BufferedReader reader;
static StringTokenizer tokenizer;
/** call this method to initialize reader for InputStream */
static void init(InputStream input) {
reader = new BufferedReader(new InputStreamReader(input));
tokenizer = new StringTokenizer("");
}
/** get next word */
static String next() throws IOException {
while (!tokenizer.hasMoreTokens()) {
// TODO add check for eof if necessary
tokenizer = new StringTokenizer(reader.readLine());
}
return tokenizer.nextToken();
}
static int nextInt() throws IOException {
return Integer.parseInt(next());
}
static long nextLong() throws IOException {
return Long.parseLong(next());
}
}
public static void main(String[] args) throws IOException{
Reader.init(System.in);
int tc =Reader.nextInt();
while (tc-->0){
int n =Reader.nextInt();
pair[] arr = new pair[n];
int count=0;
for (int i=0; i<n;i++){arr[i]=new pair(i+1,Reader.nextInt());}
mergeSort(arr, 0,n-1);
//for(pair val:arr){System.out.println(val.val+" ");}
for (int i=0; i<n;i++){
int item = 2*n/arr[i].val;
int idx= ClosestToTheLeft(arr,item,0);
for (int j=idx-1;j>i;j--){
if (arr[i].val*arr[j].val==arr[i].idx+arr[j].idx){count++;}
}
}
System.out.println(count);
}
}
static class pair{
int val;
int idx;
pair(int i, int v){
this.idx=i;
this.val=v;
}
}
static class Sortbyval implements Comparator<pair> {
public int compare(pair a, pair b)
{return a.val - b.val;}
}
public static int ClosestToTheLeft(pair[] arr, int item, int i){
int low=i;
int high = arr.length-1;
int idx=-1;
while (low<=high){
int mid=low+((high-low)/2);
if (arr[mid].val<=item){
idx = Math.max(mid,idx);
low=mid+1;}
else{high=mid-1;}
}
return idx+1;
}
static void mergeSort(pair[] arr, int low, int high) {
if (low >= high) {
return;
}
int mid = (low + high) / 2;
mergeSort(arr, low, mid);
mergeSort(arr, mid + 1, high);
merge(arr, low, high);
}
static void merge(pair[] arr, int low, int high) {
int mid = low + ((high - low) / 2);
int n1 = mid - low + 1;
int n2 = high - mid;
pair[] temp1 = new pair[n1];
pair[] temp2 = new pair[n2];
for (int i = 0; i < n1; i++) {
temp1[i] = arr[low + i];
}
long sum = 0;
for (int i = 0; i < n2; i++) {
temp2[i] = arr[mid + 1 + i];
}
int i = 0;
int j = 0;
int k = low;
while (i < n1 && j < n2) {
if (temp1[i].val <= temp2[j].val) {
arr[k] = temp1[i];
i++;
} else {
arr[k] = temp2[j];
j++;
}
k++;
}
while (i < n1) {
arr[k] = temp1[i];
i++;
k++;
}
while (j < n2) {
arr[k] = temp2[j];
j++;
k++;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | ac95ed0acff7e4b0fdc8b61b9a728bd2 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class abc {
//utilities
//Main
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long a[][] =new long[n][2];
for (int i = 0; i <n ; i++) {
a[i][0]=sc.nextInt();
a[i][1]=i+1;
}
Arrays.sort(a,(c,d)-> (int) (c[0]-d[0]));
long ans=0;
for(int i=0;i<n;i++) {
for (int j = i + 1; j < n; j++) {
if (a[i][0] *(int) a[j][0] >= 2 * n) break;
else {
if (a[i][0] *(int) a[j][0] == a[i][1] +a[j][1])
ans++;
}
}
}
System.out.println(ans);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | d9cb99a01e047b6265b48ab5f443bb47 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef {
public static void main(String[] args) throws java.lang.Exception {
FastReader in = new FastReader(System.in);
StringBuilder sb = new StringBuilder();
int t = 1;
t = in.nextInt();
while (t > 0) {
--t;
int n = in.nextInt();
long arr[] = new long[n];
for(int i = 0;i<n;i++)
arr[i] = in.nextLong();
int count = 0;
for(int i = 0;i<n;i++)
{
int k = 1;
while(arr[i]*k-(i+1)<=n && k<=2*n)
{
long x = arr[i]*k-(i+1)-1;
if(x>= 0 && x<n && arr[(int)x] == k)
{
if(x>i)
++count;
}
++k;
}
}
sb.append(count+"\n");
}
System.out.print(sb);
}
static long gcd(long a, long b) {
if (a == 0)
return b;
else
return gcd(b % a, a);
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
static void sort(long[] a) {
ArrayList<Long> l = new ArrayList<>();
for (long i : a)
l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++)
a[i] = l.get(i);
}
}
class FastReader {
byte[] buf = new byte[2048];
int index, total;
InputStream in;
FastReader(InputStream is) {
in = is;
}
int scan() throws IOException {
if (index >= total) {
index = 0;
total = in.read(buf);
if (total <= 0) {
return -1;
}
}
return buf[index++];
}
String next() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan())
;
StringBuilder sb = new StringBuilder();
for (; c > 32; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
String nextLine() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan())
;
StringBuilder sb = new StringBuilder();
for (; c != 10 && c != 13; c = scan()) {
sb.append((char) c);
}
return sb.toString();
}
char nextChar() throws IOException {
int c;
for (c = scan(); c <= 32; c = scan())
;
return (char) c;
}
int nextInt() throws IOException {
int c, val = 0;
for (c = scan(); c <= 32; c = scan())
;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
long nextLong() throws IOException {
int c;
long val = 0;
for (c = scan(); c <= 32; c = scan())
;
boolean neg = c == '-';
if (c == '-' || c == '+') {
c = scan();
}
for (; c >= '0' && c <= '9'; c = scan()) {
val = (val << 3) + (val << 1) + (c & 15);
}
return neg ? -val : val;
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | f27eddb05cc4cf91c846d0935ca6e821 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Collections;
import java.util.StringTokenizer;
public class contest728 {
public static void main(String[] args) {
FastScanner fs = new FastScanner();
int T = 1;
T=fs.nextInt();
for (int tt = 0; tt < T; tt++) {
int n= fs.nextInt();
long arr[] = new long[n+1];
for(int i=0;i<n;i++){
arr[i+1] =fs.nextLong();
}
int cnt=0;
for(int i=1;i<=n;i++){
long v= arr[i];
for(int j=(int)(arr[i]-i);j<=n;j+=(int)v){
if(j<=i){continue;}
if(arr[j]==((i+j)/arr[i])){
cnt+=1;
}
}
}
System.out.println(cnt);
}
}
static void sort(int[] a) {
ArrayList<Integer> l = new ArrayList<>();
for (int i : a) l.add(i);
Collections.sort(l);
for (int i = 0; i < a.length; i++) a[i] = l.get(i);
}
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | fb88df052d22bf73e51e944ca1f140f4 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.HashSet;
import java.util.Random;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
public final class CFPS {
static FastReader fr = new FastReader();
static PrintWriter out = new PrintWriter(System.out);
static final int gigamod = 1000000007;
static final int mod = 998244353;
static int t = 1;
static double epsilon = 0.0000000001;
static boolean[] isPrime;
static int[] smallestFactorOf;
static final int UP = 0, LEFT = 1, DOWN = 2, RIGHT = 3;
static long cmp;
@SuppressWarnings({"unused"})
public static void main(String[] args) throws Exception {
t = fr.nextInt();
OUTER:
for (int tc = 0; tc < t; tc++) {
int n = fr.nextInt();
int[] arr = fr.nextIntArray(n);
// a[i].a[j] = i + j
// --
// once a[i].a[j] exceeds 2*n, we don't need to care
int[] valIdxs = new int[2 * n + 1];
Arrays.fill(valIdxs, -1);
for (int i = 0; i < n; i++)
valIdxs[arr[i]] = i;
int ans = 0;
for (int ival = 1; ival < 2 * n + 1; ival++) {
int ii = valIdxs[ival] + 1;
if (ii == 0) continue;
for (int jval = ival + 1; jval < 2 * n + 1; jval++) {
int jj = valIdxs[jval] + 1;
if (jj == 0) continue;
if (ival * (long) jval > 2 * n)
break;
if (ival * jval == ii + jj)
ans++;
}
}
out.println(ans);
}
out.close();
}
static void compute_automaton(String s, int[][] aut) {
s += '#';
int n = s.length();
int[] pi = prefix_function(s.toCharArray());
for (int i = 0; i < n; i++) {
for (int c = 0; c < 26; c++) {
int j = i;
while (j > 0 && 'A' + c != s.charAt(j))
j = pi[j-1];
if ('A' + c == s.charAt(j))
j++;
aut[i][c] = j;
}
}
}
static void timeDFS(int current, int from, UGraph ug,
int[] time, int[] tIn, int[] tOut) {
tIn[current] = ++time[0];
for (int adj : ug.adj(current))
if (adj != from)
timeDFS(adj, current, ug, time, tIn, tOut);
tOut[current] = ++time[0];
}
static class Pair implements Comparable<Pair> {
int first, second;
int idx;
Pair() { first = second = 0; }
Pair (int ff, int ss, int ii) {
first = ff;
second = ss;
idx = ii;
}
Pair (int ff, int ss) {
first = ff;
second = ss;
idx = -1;
}
public int compareTo(Pair that) {
cmp = first - that.first;
if (cmp == 0)
cmp = second - that.second;
return (int) cmp;
}
}
static boolean areCollinear(long x1, long y1, long x2, long y2, long x3, long y3) {
// we will check if c3 lies on line through (c1, c2)
long a = x1 * (y2 - y3) + x2 * (y3 - y1) + x3 * (y1 - y2);
return a == 0;
}
static int[] treeDiameter(UGraph ug) {
int n = ug.V();
int farthest = -1;
int[] distTo = new int[n];
diamDFS(0, -1, 0, ug, distTo);
int maxDist = -1;
for (int i = 0; i < n; i++)
if (maxDist < distTo[i]) {
maxDist = distTo[i];
farthest = i;
}
distTo = new int[n + 1];
diamDFS(farthest, -1, 0, ug, distTo);
distTo[n] = farthest;
return distTo;
}
static void diamDFS(int current, int from, int dist, UGraph ug, int[] distTo) {
distTo[current] = dist;
for (int adj : ug.adj(current))
if (adj != from)
diamDFS(adj, current, dist + 1, ug, distTo);
}
static class TreeDistFinder {
UGraph ug;
int n;
int[] depthOf;
LCA lca;
TreeDistFinder(UGraph ug) {
this.ug = ug;
n = ug.V();
depthOf = new int[n];
depthCalc(0, -1, ug, 0, depthOf);
lca = new LCA(ug, 0);
}
TreeDistFinder(UGraph ug, int a) {
this.ug = ug;
n = ug.V();
depthOf = new int[n];
depthCalc(a, -1, ug, 0, depthOf);
lca = new LCA(ug, a);
}
private void depthCalc(int current, int from, UGraph ug, int depth, int[] depthOf) {
depthOf[current] = depth;
for (int adj : ug.adj(current))
if (adj != from)
depthCalc(adj, current, ug, depth + 1, depthOf);
}
public int dist(int a, int b) {
int lc = lca.lca(a, b);
return (depthOf[a] - depthOf[lc] + depthOf[b] - depthOf[lc]);
}
}
public static long[][] GCDSparseTable(long[] a)
{
int n = a.length;
int b = 32-Integer.numberOfLeadingZeros(n);
long[][] ret = new long[b][];
for(int i = 0, l = 1;i < b;i++, l*=2) {
if(i == 0) {
ret[i] = a;
} else {
ret[i] = new long[n-l+1];
for(int j = 0;j < n-l+1;j++) {
ret[i][j] = gcd(ret[i-1][j], ret[i-1][j+l/2]);
}
}
}
return ret;
}
public static long sparseRangeGCDQ(long[][] table, int l, int r)
{
// [a,b)
if(l > r)return 1;
// 1:0, 2:1, 3:1, 4:2, 5:2, 6:2, 7:2, 8:3
int t = 31-Integer.numberOfLeadingZeros(r-l);
return gcd(table[t][l], table[t][r-(1<<t)]);
}
static class Trie {
TrieNode root;
Trie(char[][] strings) {
root = new TrieNode('A', false);
construct(root, strings);
}
public Stack<String> set(TrieNode root) {
Stack<String> set = new Stack<>();
StringBuilder sb = new StringBuilder();
for (TrieNode next : root.next)
collect(sb, next, set);
return set;
}
private void collect(StringBuilder sb, TrieNode node, Stack<String> set) {
if (node == null) return;
sb.append(node.character);
if (node.isTerminal)
set.add(sb.toString());
for (TrieNode next : node.next)
collect(sb, next, set);
if (sb.length() > 0)
sb.setLength(sb.length() - 1);
}
private void construct(TrieNode root, char[][] strings) {
// we have to construct the Trie
for (char[] string : strings) {
if (string.length == 0) continue;
root.next[string[0] - 'a'] = put(root.next[string[0] - 'a'], string, 0);
if (root.next[string[0] - 'a'] != null)
root.isLeaf = false;
}
}
private TrieNode put(TrieNode node, char[] string, int idx) {
boolean isTerminal = (idx == string.length - 1);
if (node == null) node = new TrieNode(string[idx], isTerminal);
node.character = string[idx];
node.isTerminal |= isTerminal;
if (!isTerminal) {
node.isLeaf = false;
node.next[string[idx + 1] - 'a'] = put(node.next[string[idx + 1] - 'a'], string, idx + 1);
}
return node;
}
class TrieNode {
char character;
TrieNode[] next;
boolean isTerminal, isLeaf;
boolean canWin, canLose;
TrieNode(char c, boolean isTerminallll) {
character = c;
isTerminal = isTerminallll;
next = new TrieNode[26];
isLeaf = true;
}
}
}
static class Edge implements Comparable<Edge> {
int from, to;
long weight;
int id;
// int hash;
Edge(int fro, int t, long wt, int i) {
from = fro;
to = t;
id = i;
weight = wt;
// hash = Objects.hash(from, to, weight);
}
/*public int hashCode() {
return hash;
}*/
public int compareTo(Edge that) {
return Long.compare(this.id, that.id);
}
}
public static long[][] minSparseTable(long[] a)
{
int n = a.length;
int b = 32-Integer.numberOfLeadingZeros(n);
long[][] ret = new long[b][];
for(int i = 0, l = 1;i < b;i++, l*=2) {
if(i == 0) {
ret[i] = a;
}else {
ret[i] = new long[n-l+1];
for(int j = 0;j < n-l+1;j++) {
ret[i][j] = Math.min(ret[i-1][j], ret[i-1][j+l/2]);
}
}
}
return ret;
}
public static long sparseRangeMinQ(long[][] table, int l, int r)
{
// [a,b)
if(l >= r)return Integer.MAX_VALUE;
// 1:0, 2:1, 3:1, 4:2, 5:2, 6:2, 7:2, 8:3
int t = 31-Integer.numberOfLeadingZeros(r-l);
return Math.min(table[t][l], table[t][r-(1<<t)]);
}
public static long[][] maxSparseTable(long[] a)
{
int n = a.length;
int b = 32-Integer.numberOfLeadingZeros(n);
long[][] ret = new long[b][];
for(int i = 0, l = 1;i < b;i++, l*=2) {
if(i == 0) {
ret[i] = a;
}else {
ret[i] = new long[n-l+1];
for(int j = 0;j < n-l+1;j++) {
ret[i][j] = Math.max(ret[i-1][j], ret[i-1][j+l/2]);
}
}
}
return ret;
}
public static long sparseRangeMaxQ(long[][] table, int l, int r)
{
// [a,b)
if(l >= r)return Integer.MIN_VALUE;
// 1:0, 2:1, 3:1, 4:2, 5:2, 6:2, 7:2, 8:3
int t = 31-Integer.numberOfLeadingZeros(r-l);
return Math.max(table[t][l], table[t][r-(1<<t)]);
}
static class LCA {
int[] height, first, segtree;
ArrayList<Integer> euler;
boolean[] visited;
int n;
LCA(UGraph ug, int root) {
n = ug.V();
height = new int[n];
first = new int[n];
euler = new ArrayList<>();
visited = new boolean[n];
dfs(ug, root, 0);
int m = euler.size();
segtree = new int[m * 4];
build(1, 0, m - 1);
}
void dfs(UGraph ug, int node, int h) {
visited[node] = true;
height[node] = h;
first[node] = euler.size();
euler.add(node);
for (int adj : ug.adj(node)) {
if (!visited[adj]) {
dfs(ug, adj, h + 1);
euler.add(node);
}
}
}
void build(int node, int b, int e) {
if (b == e) {
segtree[node] = euler.get(b);
} else {
int mid = (b + e) / 2;
build(node << 1, b, mid);
build(node << 1 | 1, mid + 1, e);
int l = segtree[node << 1], r = segtree[node << 1 | 1];
segtree[node] = (height[l] < height[r]) ? l : r;
}
}
int query(int node, int b, int e, int L, int R) {
if (b > R || e < L)
return -1;
if (b >= L && e <= R)
return segtree[node];
int mid = (b + e) >> 1;
int left = query(node << 1, b, mid, L, R);
int right = query(node << 1 | 1, mid + 1, e, L, R);
if (left == -1) return right;
if (right == -1) return left;
return height[left] < height[right] ? left : right;
}
int lca(int u, int v) {
int left = first[u], right = first[v];
if (left > right) {
int temp = left;
left = right;
right = temp;
}
return query(1, 0, euler.size() - 1, left, right);
}
}
static class FenwickTree {
long[] array; // 1-indexed array, In this array We save cumulative information to perform efficient range queries and updates
public FenwickTree(int size) {
array = new long[size + 1];
}
public long rsq(int ind) {
assert ind > 0;
long sum = 0;
while (ind > 0) {
sum += array[ind];
//Extracting the portion up to the first significant one of the binary representation of 'ind' and decrementing ind by that number
ind -= ind & (-ind);
}
return sum;
}
public long rsq(int a, int b) {
assert b >= a && a > 0 && b > 0;
return rsq(b) - rsq(a - 1);
}
public void update(int ind, long value) {
assert ind > 0;
while (ind < array.length) {
array[ind] += value;
//Extracting the portion up to the first significant one of the binary representation of 'ind' and incrementing ind by that number
ind += ind & (-ind);
}
}
public int size() {
return array.length - 1;
}
}
static class Point implements Comparable<Point> {
long x;
long y;
long z;
long id;
// private int hashCode;
Point() {
x = z = y = 0;
// this.hashCode = Objects.hash(x, y, cost);
}
Point(Point p) {
this.x = p.x;
this.y = p.y;
this.z = p.z;
this.id = p.id;
// this.hashCode = Objects.hash(x, y, cost);
}
Point(long x, long y, long z, long id) {
this.x = x;
this.y = y;
this.z = z;
this.id = id;
// this.hashCode = Objects.hash(x, y, id);
}
Point(long a, long b) {
this.x = a;
this.y = b;
this.z = 0;
// this.hashCode = Objects.hash(a, b);
}
Point(long x, long y, long id) {
this.x = x;
this.y = y;
this.id = id;
}
@Override
public int compareTo(Point o) {
if (this.x < o.x)
return -1;
if (this.x > o.x)
return 1;
if (this.y < o.y)
return -1;
if (this.y > o.y)
return 1;
if (this.z < o.z)
return -1;
if (this.z > o.z)
return 1;
return 0;
}
@Override
public boolean equals(Object that) {
return this.compareTo((Point) that) == 0;
}
}
static class BinaryLift {
// FUNCTIONS: k-th ancestor and LCA in log(n)
int[] parentOf;
int maxJmpPow;
int[][] binAncestorOf;
int n;
int[] lvlOf;
// How this works?
// a. For every node, we store the b-ancestor for b in {1, 2, 4, 8, .. log(n)}.
// b. When we need k-ancestor, we represent 'k' in binary and for each set bit, we
// lift level in the tree.
public BinaryLift(UGraph tree) {
n = tree.V();
maxJmpPow = logk(n, 2) + 1;
parentOf = new int[n];
binAncestorOf = new int[n][maxJmpPow];
lvlOf = new int[n];
for (int i = 0; i < n; i++)
Arrays.fill(binAncestorOf[i], -1);
parentConstruct(0, -1, tree, 0);
binConstruct();
}
// TODO: Implement lvlOf[] initialization
public BinaryLift(int[] parentOf) {
this.parentOf = parentOf;
n = parentOf.length;
maxJmpPow = logk(n, 2) + 1;
binAncestorOf = new int[n][maxJmpPow];
lvlOf = new int[n];
for (int i = 0; i < n; i++)
Arrays.fill(binAncestorOf[i], -1);
UGraph tree = new UGraph(n);
for (int i = 1; i < n; i++)
tree.addEdge(i, parentOf[i]);
binConstruct();
parentConstruct(0, -1, tree, 0);
}
private void parentConstruct(int current, int from, UGraph tree, int depth) {
parentOf[current] = from;
lvlOf[current] = depth;
for (int adj : tree.adj(current))
if (adj != from)
parentConstruct(adj, current, tree, depth + 1);
}
private void binConstruct() {
for (int node = 0; node < n; node++)
for (int lvl = 0; lvl < maxJmpPow; lvl++)
binConstruct(node, lvl);
}
private int binConstruct(int node, int lvl) {
if (node < 0)
return -1;
if (lvl == 0)
return binAncestorOf[node][lvl] = parentOf[node];
if (node == 0)
return binAncestorOf[node][lvl] = -1;
if (binAncestorOf[node][lvl] != -1)
return binAncestorOf[node][lvl];
return binAncestorOf[node][lvl] = binConstruct(binConstruct(node, lvl - 1), lvl - 1);
}
// return ancestor which is 'k' levels above this one
public int ancestor(int node, int k) {
if (node < 0)
return -1;
if (node == 0)
if (k == 0) return node;
else return -1;
if (k > (1 << maxJmpPow) - 1)
return -1;
if (k == 0)
return node;
int ancestor = node;
int highestBit = Integer.highestOneBit(k);
while (k > 0 && ancestor != -1) {
ancestor = binAncestorOf[ancestor][logk(highestBit, 2)];
k -= highestBit;
highestBit = Integer.highestOneBit(k);
}
return ancestor;
}
public int lca(int u, int v) {
if (u == v)
return u;
// The invariant will be that 'u' is below 'v' initially.
if (lvlOf[u] < lvlOf[v]) {
int temp = u;
u = v;
v = temp;
}
// Equalizing the levels.
u = ancestor(u, lvlOf[u] - lvlOf[v]);
if (u == v)
return u;
// We will now raise level by largest fitting power of two until possible.
for (int power = maxJmpPow - 1; power > -1; power--)
if (binAncestorOf[u][power] != binAncestorOf[v][power]) {
u = binAncestorOf[u][power];
v = binAncestorOf[v][power];
}
return ancestor(u, 1);
}
}
static class DFSTree {
// NOTE: The thing is made keeping in mind that the whole
// input graph is connected.
UGraph tree;
UGraph backUG;
int hasBridge;
int n;
Edge backEdge;
DFSTree(UGraph ug) {
this.n = ug.V();
tree = new UGraph(n);
hasBridge = -1;
backUG = new UGraph(n);
treeCalc(0, -1, new boolean[n], ug);
}
private void treeCalc(int current, int from, boolean[] marked, UGraph ug) {
if (marked[current]) {
// This is a backEdge.
backUG.addEdge(from, current);
backEdge = new Edge(from, current, 1, 0);
return;
}
if (from != -1)
tree.addEdge(from, current);
marked[current] = true;
for (int adj : ug.adj(current))
if (adj != from)
treeCalc(adj, current, marked, ug);
}
public boolean hasBridge() {
if (hasBridge != -1)
return (hasBridge == 1);
// We have to determine the bridge.
bridgeFinder();
return (hasBridge == 1);
}
int[] levelOf;
int[] dp;
private void bridgeFinder() {
// Finding the level of each node.
levelOf = new int[n];
levelDFS(0, -1, 0);
// Applying DP solution.
// dp[i] -> Highest level reachable from subtree of 'i' using
// some backEdge.
dp = new int[n];
Arrays.fill(dp, Integer.MAX_VALUE / 100);
dpDFS(0, -1);
// Now, we will check each edge and determine whether its a
// bridge.
for (int i = 0; i < n; i++)
for (int adj : tree.adj(i)) {
// (i -> adj) is the edge.
if (dp[adj] > levelOf[i])
hasBridge = 1;
}
if (hasBridge != 1)
hasBridge = 0;
}
private void levelDFS(int current, int from, int lvl) {
levelOf[current] = lvl;
for (int adj : tree.adj(current))
if (adj != from)
levelDFS(adj, current, lvl + 1);
}
private int dpDFS(int current, int from) {
dp[current] = levelOf[current];
for (int back : backUG.adj(current))
dp[current] = Math.min(dp[current], levelOf[back]);
for (int adj : tree.adj(current))
if (adj != from)
dp[current] = Math.min(dp[current], dpDFS(adj, current));
return dp[current];
}
}
static class UnionFind {
// Uses weighted quick-union with path compression.
private int[] parent; // parent[i] = parent of i
private int[] size; // size[i] = number of sites in tree rooted at i
// Note: not necessarily correct if i is not a root node
private int count; // number of components
public UnionFind(int n) {
count = n;
parent = new int[n];
size = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
// Number of connected components.
public int count() {
return count;
}
// Find the root of p.
public int find(int p) {
while (p != parent[p])
p = parent[p];
return p;
}
public boolean connected(int p, int q) {
return find(p) == find(q);
}
public int numConnectedTo(int node) {
return size[find(node)];
}
// Weighted union.
public void union(int p, int q) {
int rootP = find(p);
int rootQ = find(q);
if (rootP == rootQ) return;
// make smaller root point to larger one
if (size[rootP] < size[rootQ]) {
parent[rootP] = rootQ;
size[rootQ] += size[rootP];
}
else {
parent[rootQ] = rootP;
size[rootP] += size[rootQ];
}
count--;
}
public static int[] connectedComponents(UnionFind uf) {
// We can do this in nlogn.
int n = uf.size.length;
int[] compoColors = new int[n];
for (int i = 0; i < n; i++)
compoColors[i] = uf.find(i);
HashMap<Integer, Integer> oldToNew = new HashMap<>();
int newCtr = 0;
for (int i = 0; i < n; i++) {
int thisOldColor = compoColors[i];
Integer thisNewColor = oldToNew.get(thisOldColor);
if (thisNewColor == null)
thisNewColor = newCtr++;
oldToNew.put(thisOldColor, thisNewColor);
compoColors[i] = thisNewColor;
}
return compoColors;
}
}
static class UGraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
@SuppressWarnings("unchecked")
public UGraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
adj[to].add(from);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int degree(int v) {
return adj[v].size();
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static void dfsMark(int current, boolean[] marked, UGraph g) {
if (marked[current]) return;
marked[current] = true;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, marked, g);
}
public static void dfsMark(int current, int from, long[] distTo, boolean[] marked, UGraph g, ArrayList<Integer> endPoints) {
if (marked[current]) return;
marked[current] = true;
if (from != -1)
distTo[current] = distTo[from] + 1;
HashSet<Integer> adj = g.adj(current);
int alreadyMarkedCtr = 0;
for (int adjc : adj) {
if (marked[adjc]) alreadyMarkedCtr++;
dfsMark(adjc, current, distTo, marked, g, endPoints);
}
if (alreadyMarkedCtr == adj.size())
endPoints.add(current);
}
public static void bfsOrder(int current, UGraph g) {
}
public static void dfsMark(int current, int[] colorIds, int color, UGraph g) {
if (colorIds[current] != -1) return;
colorIds[current] = color;
Iterable<Integer> adj = g.adj(current);
for (int adjc : adj)
dfsMark(adjc, colorIds, color, g);
}
public static int[] connectedComponents(UGraph g) {
int n = g.V();
int[] componentId = new int[n];
Arrays.fill(componentId, -1);
int colorCtr = 0;
for (int i = 0; i < n; i++) {
if (componentId[i] != -1) continue;
dfsMark(i, componentId, colorCtr, g);
colorCtr++;
}
return componentId;
}
public static boolean hasCycle(UGraph ug) {
int n = ug.V();
boolean[] marked = new boolean[n];
boolean[] hasCycleFirst = new boolean[1];
for (int i = 0; i < n; i++) {
if (marked[i]) continue;
hcDfsMark(i, ug, marked, hasCycleFirst, -1);
}
return hasCycleFirst[0];
}
// Helper for hasCycle.
private static void hcDfsMark(int current, UGraph ug, boolean[] marked, boolean[] hasCycleFirst, int parent) {
if (marked[current]) return;
if (hasCycleFirst[0]) return;
marked[current] = true;
HashSet<Integer> adjc = ug.adj(current);
for (int adj : adjc) {
if (marked[adj] && adj != parent && parent != -1) {
hasCycleFirst[0] = true;
return;
}
hcDfsMark(adj, ug, marked, hasCycleFirst, current);
}
}
}
static class Digraph {
// Adjacency list.
private HashSet<Integer>[] adj;
private static final String NEWLINE = "\n";
private int E;
@SuppressWarnings("unchecked")
public Digraph(int V) {
adj = (HashSet<Integer>[]) new HashSet[V];
E = 0;
for (int i = 0; i < V; i++)
adj[i] = new HashSet<Integer>();
}
public void addEdge(int from, int to) {
if (adj[from].contains(to)) return;
E++;
adj[from].add(to);
}
public HashSet<Integer> adj(int from) {
return adj[from];
}
public int V() {
return adj.length;
}
public int E() {
return E;
}
public Digraph reversed() {
Digraph dg = new Digraph(V());
for (int i = 0; i < V(); i++)
for (int adjVert : adj(i)) dg.addEdge(adjVert, i);
return dg;
}
public String toString() {
StringBuilder s = new StringBuilder();
s.append(V() + " vertices, " + E() + " edges " + NEWLINE);
for (int v = 0; v < V(); v++) {
s.append(v + ": ");
for (int w : adj[v]) {
s.append(w + " ");
}
s.append(NEWLINE);
}
return s.toString();
}
public static int[] KosarajuSharirSCC(Digraph dg) {
int[] id = new int[dg.V()];
Digraph reversed = dg.reversed();
// Gotta perform topological sort on this one to get the stack.
Stack<Integer> revStack = Digraph.topologicalSort(reversed);
// Initializing id and idCtr.
id = new int[dg.V()];
int idCtr = -1;
// Creating a 'marked' array.
boolean[] marked = new boolean[dg.V()];
while (!revStack.isEmpty()) {
int vertex = revStack.pop();
if (!marked[vertex])
sccDFS(dg, vertex, marked, ++idCtr, id);
}
return id;
}
private static void sccDFS(Digraph dg, int source, boolean[] marked, int idCtr, int[] id) {
marked[source] = true;
id[source] = idCtr;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex]) sccDFS(dg, adjVertex, marked, idCtr, id);
}
public static Stack<Integer> topologicalSort(Digraph dg) {
// dg has to be a directed acyclic graph.
// We'll have to run dfs on the digraph and push the deepest nodes on stack first.
// We'll need a Stack<Integer> and a int[] marked.
Stack<Integer> topologicalStack = new Stack<Integer>();
boolean[] marked = new boolean[dg.V()];
// Calling dfs
for (int i = 0; i < dg.V(); i++)
if (!marked[i])
runDfs(dg, topologicalStack, marked, i);
return topologicalStack;
}
static void runDfs(Digraph dg, Stack<Integer> topologicalStack, boolean[] marked, int source) {
marked[source] = true;
for (Integer adjVertex : dg.adj(source))
if (!marked[adjVertex])
runDfs(dg, topologicalStack, marked, adjVertex);
topologicalStack.add(source);
}
}
static class FastReader {
private BufferedReader bfr;
private StringTokenizer st;
public FastReader() {
bfr = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
if (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(bfr.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
char nextChar() {
return next().toCharArray()[0];
}
String nextString() {
return next();
}
int[] nextIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++)
arr[i] = nextInt();
return arr;
}
double[] nextDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < arr.length; i++)
arr[i] = nextDouble();
return arr;
}
long[] nextLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++)
arr[i] = nextLong();
return arr;
}
int[][] nextIntGrid(int n, int m) {
int[][] grid = new int[n][m];
for (int i = 0; i < n; i++) {
for (int j = 0; j < m; j++)
grid[i][j] = fr.nextInt();
}
return grid;
}
}
@SuppressWarnings("serial")
static class CountMap<T> extends TreeMap<T, Integer>{
CountMap() {
}
CountMap(Comparator<T> cmp) {
}
CountMap(T[] arr) {
this.putCM(arr);
}
public Integer putCM(T key) {
return super.put(key, super.getOrDefault(key, 0) + 1);
}
public Integer removeCM(T key) {
int count = super.getOrDefault(key, -1);
if (count == -1) return -1;
if (count == 1)
return super.remove(key);
else
return super.put(key, count - 1);
}
public Integer getCM(T key) {
return super.getOrDefault(key, 0);
}
public void putCM(T[] arr) {
for (T l : arr)
this.putCM(l);
}
}
static long dioGCD(long a, long b, long[] x0, long[] y0) {
if (b == 0) {
x0[0] = 1;
y0[0] = 0;
return a;
}
long[] x1 = new long[1], y1 = new long[1];
long d = dioGCD(b, a % b, x1, y1);
x0[0] = y1[0];
y0[0] = x1[0] - y1[0] * (a / b);
return d;
}
static boolean diophantine(long a, long b, long c, long[] x0, long[] y0, long[] g) {
g[0] = dioGCD(Math.abs(a), Math.abs(b), x0, y0);
if (c % g[0] > 0) {
return false;
}
x0[0] *= c / g[0];
y0[0] *= c / g[0];
if (a < 0) x0[0] = -x0[0];
if (b < 0) y0[0] = -y0[0];
return true;
}
static long[][] prod(long[][] mat1, long[][] mat2) {
int n = mat1.length;
long[][] prod = new long[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
// determining prod[i][j]
// it will be the dot product of mat1[i][] and mat2[][i]
for (int k = 0; k < n; k++)
prod[i][j] += mat1[i][k] * mat2[k][j];
return prod;
}
static long[][] matExpo(long[][] mat, long power) {
int n = mat.length;
long[][] ans = new long[n][n];
if (power == 0)
return null;
if (power == 1)
return mat;
long[][] half = matExpo(mat, power / 2);
ans = prod(half, half);
if (power % 2 == 1) {
ans = prod(ans, mat);
}
return ans;
}
static int KMPNumOcc(char[] text, char[] pat) {
int n = text.length;
int m = pat.length;
char[] patPlusText = new char[n + m + 1];
for (int i = 0; i < m; i++)
patPlusText[i] = pat[i];
patPlusText[m] = '^'; // Seperator
for (int i = 0; i < n; i++)
patPlusText[m + i] = text[i];
int[] fullPi = piCalcKMP(patPlusText);
int answer = 0;
for (int i = 0; i < n + m + 1; i++)
if (fullPi[i] == m)
answer++;
return answer;
}
static int[] piCalcKMP(char[] s) {
int n = s.length;
int[] pi = new int[n];
for (int i = 1; i < n; i++) {
int j = pi[i - 1];
while (j > 0 && s[i] != s[j])
j = pi[j - 1];
if (s[i] == s[j])
j++;
pi[i] = j;
}
return pi;
}
static boolean[] prefMatchesSuff(char[] s) {
int n = s.length;
boolean[] res = new boolean[n + 1];
int[] pi = prefix_function(s);
res[0] = true;
for (int p = n; p != 0; p = pi[p])
res[p] = true;
return res;
}
static int[] prefix_function(char[] s) {
int n = s.length;
int[] pi = new int[n];
for (int i = 1; i < n; i++) {
int j = pi[i-1];
while (j > 0 && s[i] != s[j])
j = pi[j-1];
if (s[i] == s[j])
j++;
pi[i] = j;
}
return pi;
}
static long hash(long key) {
long h = Long.hashCode(key);
h ^= (h >>> 20) ^ (h >>> 12) ^ (h >>> 7) ^ (h >>> 4);
return h & (gigamod-1);
}
static int mapTo1D(int row, int col, int n, int m) {
// Maps elements in a 2D matrix serially to elements in
// a 1D array.
return row * m + col;
}
static int[] mapTo2D(int idx, int n, int m) {
// Inverse of what the one above does.
int[] rnc = new int[2];
rnc[0] = idx / m;
rnc[1] = idx % m;
return rnc;
}
static boolean[] primeGenerator(int upto) {
// Sieve of Eratosthenes:
isPrime = new boolean[upto + 1];
smallestFactorOf = new int[upto + 1];
Arrays.fill(smallestFactorOf, 1);
Arrays.fill(isPrime, true);
isPrime[1] = isPrime[0] = false;
for (long i = 2; i < upto + 1; i++)
if (isPrime[(int) i]) {
smallestFactorOf[(int) i] = (int) i;
// Mark all the multiples greater than or equal
// to the square of i to be false.
for (long j = i; j * i < upto + 1; j++) {
if (isPrime[(int) j * (int) i]) {
isPrime[(int) j * (int) i] = false;
smallestFactorOf[(int) j * (int) i] = (int) i;
}
}
}
return isPrime;
}
static HashMap<Integer, Integer> smolNumPrimeFactorization(int num) {
if (smallestFactorOf == null)
primeGenerator(num + 1);
HashMap<Integer, Integer> fnps = new HashMap<>();
while (num != 1) {
fnps.put(smallestFactorOf[num], fnps.getOrDefault(smallestFactorOf[num], 0) + 1);
num /= smallestFactorOf[num];
}
return fnps;
}
static HashMap<Long, Integer> primeFactorization(long num) {
// Returns map of factor and its power in the number.
HashMap<Long, Integer> map = new HashMap<>();
while (num % 2 == 0) {
num /= 2;
Integer pwrCnt = map.get(2L);
map.put(2L, pwrCnt != null ? pwrCnt + 1 : 1);
}
for (long i = 3; i * i <= num; i += 2) {
while (num % i == 0) {
num /= i;
Integer pwrCnt = map.get(i);
map.put(i, pwrCnt != null ? pwrCnt + 1 : 1);
}
}
// If the number is prime, we have to add it to the
// map.
if (num != 1)
map.put(num, 1);
return map;
}
static HashSet<Long> divisors(long num) {
HashSet<Long> divisors = new HashSet<Long>();
divisors.add(1L);
divisors.add(num);
for (long i = 2; i * i <= num; i++) {
if (num % i == 0) {
divisors.add(num/i);
divisors.add(i);
}
}
return divisors;
}
static void coprimeGenerator(int m, int n, ArrayList<Point> coprimes, int limit, int numCoprimes) {
if (m > limit) return;
if (m <= limit && n <= limit)
coprimes.add(new Point(m, n));
if (coprimes.size() > numCoprimes) return;
coprimeGenerator(2 * m - n, m, coprimes, limit, numCoprimes);
coprimeGenerator(2 * m + n, m, coprimes, limit, numCoprimes);
coprimeGenerator(m + 2 * n, n, coprimes, limit, numCoprimes);
}
static long nCr(long n, long r, long[] fac) { long p = gigamod; if (r == 0) return 1; return (fac[(int)n] * modInverse(fac[(int)r], p) % p * modInverse(fac[(int)n - (int)r], p) % p) % p; }
static long modInverse(long n, long p) { return power(n, p - 2, p); }
static long modDiv(long a, long b){return mod(a * power(b, mod - 2, mod), mod);}
static long power(long x, long y, long p) { long res = 1; x = x % p; while (y > 0) { if ((y & 1)==1) res = (res * x) % p; y = y >> 1; x = (x * x) % p; } return res; }
static int logk(long n, long k) { return (int)(Math.log(n) / Math.log(k)); }
static long gcd(long a, long b) { if (b == 0) { return a; } else { return gcd(b, a % b); } } static int gcd(int a, int b) { if (b == 0) { return a; } else { return gcd(b, a % b); } } static long gcd(long[] arr) { int n = arr.length; long gcd = arr[0]; for (int i = 1; i < n; i++) { gcd = gcd(gcd, arr[i]); } return gcd; } static int gcd(int[] arr) { int n = arr.length; int gcd = arr[0]; for (int i = 1; i < n; i++) { gcd = gcd(gcd, arr[i]); } return gcd; } static long lcm(long[] arr) { long lcm = arr[0]; int n = arr.length; for (int i = 1; i < n; i++) { lcm = (lcm * arr[i]) / gcd(lcm, arr[i]); } return lcm; } static long lcm(long a, long b) { return (a * b)/gcd(a, b); } static boolean less(int a, int b) { return a < b ? true : false; } static boolean isSorted(int[] a) { for (int i = 1; i < a.length; i++) { if (less(a[i], a[i - 1])) return false; } return true; } static boolean isSorted(long[] a) { for (int i = 1; i < a.length; i++) { if (a[i] < a[i - 1]) return false; } return true; } static void swap(int[] a, int i, int j) { int temp = a[i]; a[i] = a[j]; a[j] = temp; } static void swap(long[] a, int i, int j) { long temp = a[i]; a[i] = a[j]; a[j] = temp; } static void swap(double[] a, int i, int j) { double temp = a[i]; a[i] = a[j]; a[j] = temp; } static void swap(char[] a, int i, int j) { char temp = a[i]; a[i] = a[j]; a[j] = temp; }
static void sort(int[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); } static void sort(char[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); } static void sort(long[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); } static void sort(double[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); } static void reverseSort(int[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); int n = arr.length; for (int i = 0; i < n/2; i++) swap(arr, i, n - 1 - i); } static void reverseSort(char[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); int n = arr.length; for (int i = 0; i < n/2; i++) swap(arr, i, n - 1 - i); } static void reverse(char[] arr) { int n = arr.length; for (int i = 0; i < n / 2; i++) swap(arr, i, n - 1 - i); } static void reverseSort(long[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); int n = arr.length; for (int i = 0; i < n/2; i++) swap(arr, i, n - 1 - i); } static void reverseSort(double[] arr) { shuffleArray(arr, 0, arr.length - 1); Arrays.sort(arr); int n = arr.length; for (int i = 0; i < n/2; i++) swap(arr, i, n - 1 - i); }
static void shuffleArray(long[] arr, int startPos, int endPos) { Random rnd = new Random(); for (int i = startPos; i < endPos; ++i) { long tmp = arr[i]; int randomPos = i + rnd.nextInt(endPos - i); arr[i] = arr[randomPos]; arr[randomPos] = tmp; } } static void shuffleArray(int[] arr, int startPos, int endPos) { Random rnd = new Random(); for (int i = startPos; i < endPos; ++i) { int tmp = arr[i]; int randomPos = i + rnd.nextInt(endPos - i); arr[i] = arr[randomPos]; arr[randomPos] = tmp; } }
static void shuffleArray(double[] arr, int startPos, int endPos) { Random rnd = new Random(); for (int i = startPos; i < endPos; ++i) { double tmp = arr[i]; int randomPos = i + rnd.nextInt(endPos - i); arr[i] = arr[randomPos]; arr[randomPos] = tmp; } }
static void shuffleArray(char[] arr, int startPos, int endPos) { Random rnd = new Random(); for (int i = startPos; i < endPos; ++i) { char tmp = arr[i]; int randomPos = i + rnd.nextInt(endPos - i); arr[i] = arr[randomPos]; arr[randomPos] = tmp; } }
static boolean isPrime(long n) {if (n<=1)return false;if(n<=3)return true;if(n%2==0||n%3==0)return false;for(long i=5;i*i<=n;i=i+6)if(n%i==0||n%(i+2)==0)return false;return true;}
static String toString(int[] dp){StringBuilder sb=new StringBuilder();for(int i=0;i<dp.length;i++)sb.append(dp[i]+" ");return sb.toString();}
static String toString(boolean[] dp){StringBuilder sb=new StringBuilder();for(int i=0;i<dp.length;i++)sb.append(dp[i]+" ");return sb.toString();}
static String toString(long[] dp){StringBuilder sb=new StringBuilder();for(int i=0;i<dp.length;i++)sb.append(dp[i]+" ");return sb.toString();}
static String toString(char[] dp){StringBuilder sb=new StringBuilder();for(int i=0;i<dp.length;i++)sb.append(dp[i]+"");return sb.toString();}
static String toString(int[][] dp){StringBuilder sb=new StringBuilder();for(int i=0;i<dp.length;i++){for(int j=0;j<dp[i].length;j++){sb.append(dp[i][j]+" ");}sb.append('\n');}return sb.toString();}
static String toString(long[][] dp){StringBuilder sb=new StringBuilder();for(int i=0;i<dp.length;i++){for(int j=0;j<dp[i].length;j++) {sb.append(dp[i][j]+" ");}sb.append('\n');}return sb.toString();}
static String toString(double[][] dp){StringBuilder sb=new StringBuilder();for(int i = 0;i<dp.length;i++){for(int j = 0;j<dp[i].length;j++){sb.append(dp[i][j]+" ");}sb.append('\n');}return sb.toString();}
static String toString(char[][] dp){StringBuilder sb = new StringBuilder();for(int i = 0;i<dp.length;i++){for(int j = 0;j<dp[i].length;j++){sb.append(dp[i][j]+"");}sb.append('\n');}return sb.toString();}
static long mod(long a, long m){return(a%m+1000000L*m)%m;}
}
// NOTES:
// ASCII VALUE OF 'A': 65
// ASCII VALUE OF 'a': 97
// Range of long: 9 * 10^18
// ASCII VALUE OF '0': 48
// Primes upto 'n' can be given by (n / (logn)). | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | ba3a75c52d0e104a14e02035708663ac | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class ProblemE {
public static void main(String[] args) throws IOException {
Reader io = new Reader();
PrintWriter pw = new PrintWriter(System.out);
int t = io.nextInt();
while(t-- > 0) {
int n = io.nextInt();
long[] arr = new long[n + 1];
for (int i = 1; i < arr.length; i++) {
arr[i] = io.nextInt();
}
int ans = 0;
for(int i = 1; i <= n; i++) {
for(int j = (int)arr[i] - i; j <= n; j+=arr[i]) {
if(j >= 0) {
if((arr[i] * arr[j] == i + j) && i < j) {
ans++;
}
}
}
}
System.out.println(ans);
}
io.close();
pw.close();
}
static int INF = Integer.MAX_VALUE;
static long LINF = Long.MAX_VALUE;
public static class Pair<A, B> {
A first;
B second;
// Constructor
public Pair(A first, B second)
{
this.first = first;
this.second = second;
}
}
// Kattio is too slow
//BeginCodeSnip{Fast Reader}
static class Reader
{
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do
{
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
}
//EndCodeSnip{}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 8 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 21da2d8b796b661d67061b1826079d41 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class PleasantPair {
public static void main(String[] args) {
FastReader s = new FastReader();
int t = s.nextInt();
try{
while (t>0)
{
int count=0;
int n=s.nextInt();
long [] a= new long[n+1];
for(int i=1;i<=n;i++)
{
a[i]= s.nextLong();
}
for(int i=1;i<=n;i++)
{
for(int j = (int) a[i] - i; j <= n; j += a[i])
{
if(j>0)
{
if(i < j && (long) a[i]*a[j] ==(long) i+j)
count++;
}
}
}
System.out.println(count);
t--;
}
}
catch (Exception e)
{
}
/*
FastReader fr = new FastReader();
int numCas = fr.nextInt();
long[] arr;
while(numCas-- > 0) {
int length = fr.nextInt();
arr = new long[length + 1];
for(int i = 1; i <= length; i++) {
arr[i] = fr.nextLong();
}
count(arr);
}
*/
}
/*
public static void count(long[] arr) {
int count = 0;
for(int i = 1; i <= arr.length; i++) {
for(int j = (int)arr[i] - i; j <= arr.length; j+= arr[i]) {
if(j>0) {
if((long)(arr[i] * arr[j]) == (long)i + j) {
count++;
}
}
}
}
System.out.println(count);
}
*/
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
public FastReader(File f) throws IOException {
br = new BufferedReader(new FileReader(f));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
int [] readintarray(int n) {
int res [] = new int [n];
for(int i = 0; i<n; i++)res[i] = nextInt();
return res;
}
long [] readlongarray(int n) {
long res [] = new long [n];
for(int i = 0; i<n; i++)res[i] = nextLong();
return res;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | f6b68d795d58dd164b3058f92889a79d | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;import java.lang.*;import java.util.*;
// number of prime numbers less then or equal to x are --> x/ln(x)
public class b {static class FastScanner {InputStreamReader is;BufferedReader br;StringTokenizer st;
public FastScanner() {is = new InputStreamReader(System.in);br = new BufferedReader(is);}
String next() throws Exception {while (st == null || !st.hasMoreElements())st = new StringTokenizer(br.readLine());
return st.nextToken();}int nextInt() throws Exception {return Integer.parseInt(next());}long nextLong() throws Exception {
return Long.parseLong(next());}int[] readArray(int num) throws Exception {int arr[]=new int[num];
for(int i=0;i<num;i++)arr[i]=nextInt();return arr;}String nextLine() throws Exception {return br.readLine();
}} public static boolean power_of_two(int a){if((a&(a-1))==0){ return true;}return false;}
static boolean PS(double x){if (x >= 0) {double i= Math.sqrt(x);if(i%1!=0){
return false;}return ((i * i) == x);}return false;}public static int[] ia(int n){int ar[]=new int[n];
return ar;}public static long[] la(int n)
{long ar[]=new long[n];return ar;}
public static void print(int ans,int t){System.out.println("Case"+" "+"#"+t+":"+" "+ans);}
static long mod=1000000007;
static int max=Integer.MIN_VALUE;
static int min=Integer.MAX_VALUE;
public static void main(String args[]) throws java.lang.Exception
{
FastScanner sc=new FastScanner();
int t=sc.nextInt();
while(t-->0)
{int n=sc.nextInt();int ar[]=new int[n+1];
for(int i=1;i<=n;i++)
{
ar[i]=sc.nextInt();
}
long ans=0;
for(int i=1;i<=n;i++)
{
int st=ar[i]-(i);
for(int j=st;j<=n;j+=ar[i])
{
if(j>0)
{
if((long)ar[i]*ar[j]==i+j && j>i)
{
ans++;
}
}
}
}
System.out.println(ans);
}}} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | a34a27d2dc32b0767c95cad81ede6bef | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;import java.lang.*;import java.util.*;
// number of prime numbers less then or equal to x are --> x/ln(x)
public class b {static class FastScanner {InputStreamReader is;BufferedReader br;StringTokenizer st;
public FastScanner() {is = new InputStreamReader(System.in);br = new BufferedReader(is);}
String next() throws Exception {while (st == null || !st.hasMoreElements())st = new StringTokenizer(br.readLine());
return st.nextToken();}int nextInt() throws Exception {return Integer.parseInt(next());}long nextLong() throws Exception {
return Long.parseLong(next());}int[] readArray(int num) throws Exception {int arr[]=new int[num];
for(int i=0;i<num;i++)arr[i]=nextInt();return arr;}String nextLine() throws Exception {return br.readLine();
}} public static boolean power_of_two(int a){if((a&(a-1))==0){ return true;}return false;}
static boolean PS(double x){if (x >= 0) {double i= Math.sqrt(x);if(i%1!=0){
return false;}return ((i * i) == x);}return false;}public static int[] ia(int n){int ar[]=new int[n];
return ar;}public static long[] la(int n)
{long ar[]=new long[n];return ar;}
public static void print(int ans,int t){System.out.println("Case"+" "+"#"+t+":"+" "+ans);}
static long mod=1000000007;
static int max=Integer.MIN_VALUE;
static int min=Integer.MAX_VALUE;
public static void main(String args[]) throws java.lang.Exception
{
FastScanner sc=new FastScanner();
int t=sc.nextInt();
while(t-->0)
{int n=sc.nextInt();int ar[]=new int[n+1];
for(int i=1;i<=n;i++)
{
ar[i]=sc.nextInt();
}
long ans=0;
for(int i=1;i<=n;i++)
{
int st=ar[i]-(i);
for(int j=st;j<=n;j+=ar[i])
{
if(j>0)
{
if((long)ar[i]*ar[j]==i+j && j>i)
{
ans++;
}
}
}
}
System.out.println(ans);
}}} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 5782d9d8aa1b219a5fbf28cfebdeb177 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;import java.lang.*;import java.util.*;
// number of prime numbers less then or equal to x are --> x/ln(x)
public class b {static class FastScanner {InputStreamReader is;BufferedReader br;StringTokenizer st;
public FastScanner() {is = new InputStreamReader(System.in);br = new BufferedReader(is);}
String next() throws Exception {while (st == null || !st.hasMoreElements())st = new StringTokenizer(br.readLine());
return st.nextToken();}int nextInt() throws Exception {return Integer.parseInt(next());}long nextLong() throws Exception {
return Long.parseLong(next());}int[] readArray(int num) throws Exception {int arr[]=new int[num];
for(int i=0;i<num;i++)arr[i]=nextInt();return arr;}String nextLine() throws Exception {return br.readLine();
}} public static boolean power_of_two(int a){if((a&(a-1))==0){ return true;}return false;}
static boolean PS(double x){if (x >= 0) {double i= Math.sqrt(x);if(i%1!=0){
return false;}return ((i * i) == x);}return false;}public static int[] ia(int n){int ar[]=new int[n];
return ar;}public static long[] la(int n)
{long ar[]=new long[n];return ar;}
public static void print(int ans,int t){System.out.println("Case"+" "+"#"+t+":"+" "+ans);}
static long mod=1000000007;
static int max=Integer.MIN_VALUE;
static int min=Integer.MAX_VALUE;
public static void main(String args[]) throws java.lang.Exception
{
FastScanner sc=new FastScanner();
int t=sc.nextInt();
while(t-->0)
{int n=sc.nextInt();int ar[]=new int[n+1];
for(int i=1;i<=n;i++)
{
ar[i]=sc.nextInt();
}
long ans=0;
for(int i=1;i<=n;i++)
{
int st=ar[i]-(i);
for(int j=st;j<=n;j+=ar[i])
{
if(j>0)
{
if((long)ar[i]*ar[j]==i+j && j>i)
{
ans++;
}
}
}
}
System.out.println(ans);
}}} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 52c9db00e98342b62e0f41d914d00746 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Scanner;
public class edoti {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
long[] a = new long[n + 1];
for (int i = 1; i < n + 1; i++) {
a[i] = s.nextInt();
}
long count = 0;
for (int i = 1; i <= n; i++) {
for (int j = (int)a[i]-i; j <= n; j += a[i]) {
if (j > 0) {
if (a[i] * a[j] == i + j && j > i) {
count++;
}
}
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 9cf87fb1451f5b209af38a8dfaf0202b | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Scanner;
public class edoti {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
long[] a = new long[n + 1];
for (int i = 1; i < n + 1; i++) {
a[i] = s.nextInt();
}
long count = 0;
for (int i = 1; i <= n; i++) {
for (int j = (int)a[i]-i; j <= n; j += a[i]) {
if (j > 0) {
if (a[i] * a[j] == i + j && j > i) {
count++;
}
}
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 5da0e36397fcb176362ecd7b7eb02190 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Scanner;
public class edoti {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while (t-- > 0) {
int n = s.nextInt();
long[] a = new long[n + 1];
for (int i = 1; i < n + 1; i++) {
a[i] = s.nextInt();
}
int count = 0;
for (int i = 1; i <= n; i++) {
for (int j = (int) (a[i] - i); j <= n; j += a[i]) {
if (j > 0) {
if (a[i] * a[j] == i + j && j > i) {
count++;
}
}
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | ba5dcbab92e675d6a000d85ff4677ba5 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class b {
static PrintWriter out = new PrintWriter(System.out);
static void solve(int arr[], int n) {
Pair pairs[] = new Pair[n];
for (int i = 0; i < n; i++) {
pairs[i] = new Pair(arr[i], i + 1);
}
Arrays.sort(pairs);
int count = 0;
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
if (pairs[i - 1].second + pairs[j - 1].second == (long) pairs[i - 1].first * pairs[j - 1].first)
count++;
if ((long) pairs[i - 1].first * pairs[j - 1].first > 2 * n)
break;
}
}
out.println(count);
}
// static void solve(int arr[], int n) {
// int[] index = new int[2 * n + 1];
// for (int i = 0; i < n; i++) {
// index[arr[i]] = i + 1;
// }
// Arrays.sort(arr);
// long res = 0;
// for (int i = 0; i < n; i++) {
// for (int j = i + 1; j < n; j++) {
// if (index[arr[i]] + index[arr[j]] == (long) arr[i] * arr[j])
// res++;
// if ((long) arr[i] * arr[j] > 2 * n)
// break;
// }
// }
// out.println(res);
// }
static void sort(int[] arr) {
// shuffle
Random rand = new Random();
for (int i = 0; i < arr.length; i++) {
int j = rand.nextInt(i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
Arrays.sort(arr);
}
public static void main(String[] args) throws Exception {
FastScanner fs = new FastScanner();
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int arr[] = fs.readArray(n);
solve(arr, n);
}
out.close();
}
static class Pair implements Comparable<Pair> {
int first, second;
Pair(int f, int s) {
this.first = f;
this.second = s;
}
public int compareTo(Pair o) {
return this.first - o.first;
}
}
//////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 4f35a1a5d23df8e22c800f4a9c477ac3 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class b {
static PrintWriter out = new PrintWriter(System.out);
static void solve(int arr[], int n) {
Pair pairs[] = new Pair[n];
for (int i = 0; i < n; i++) {
pairs[i] = new Pair(arr[i], i + 1);
}
Arrays.sort(pairs);
long count = 0;
for (int i = 1; i <= n; i++) {
for (int j = i + 1; j <= n; j++) {
if (pairs[i - 1].second + pairs[j - 1].second == (long) pairs[i - 1].first * pairs[j - 1].first)
count++;
if ((long) pairs[i - 1].first * pairs[j - 1].first > 2 * n)
break;
}
}
out.println(count);
}
// static void solve(int arr[], int n) {
// int[] index = new int[2 * n + 1];
// for (int i = 0; i < n; i++) {
// index[arr[i]] = i + 1;
// }
// Arrays.sort(arr);
// long res = 0;
// for (int i = 0; i < n; i++) {
// for (int j = i + 1; j < n; j++) {
// if (index[arr[i]] + index[arr[j]] == (long) arr[i] * arr[j])
// res++;
// if ((long) arr[i] * arr[j] > 2 * n)
// break;
// }
// }
// out.println(res);
// }
static void sort(int[] arr) {
// shuffle
Random rand = new Random();
for (int i = 0; i < arr.length; i++) {
int j = rand.nextInt(i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
Arrays.sort(arr);
}
public static void main(String[] args) throws Exception {
FastScanner fs = new FastScanner();
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int arr[] = fs.readArray(n);
solve(arr, n);
}
out.close();
}
static class Pair implements Comparable<Pair> {
int first, second;
Pair(int f, int s) {
this.first = f;
this.second = s;
}
public int compareTo(Pair o) {
return this.first - o.first;
}
}
//////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 9a27073ff8aa5bf8217a713544900185 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class b {
static PrintWriter out = new PrintWriter(System.out);
// static void solve(int arr[], int n) {
// Pair pairs[] = new Pair[n];
// for (int i = 0; i < n; i++) {
// pairs[i] = new Pair(arr[i], i + 1);
// }
// Arrays.sort(pairs);
// long count = 0;
// for (int i = 1; i <= n; i++) {
// for (int j = i + 1; j <= n; j++) {
// long x = pairs[i - 1].second + pairs[j - 1].second; // i+j
// long y = pairs[i - 1].first * pairs[j - 1].first;// ai*aj
// if (y > 2 * n)
// break; // ai.aj<=2n
// if (x == y)
// count++;
// }
// }
// out.println(count);
// }
static void solve(int arr[], int n) {
int[] index = new int[2 * n + 1];
for (int i = 0; i < n; i++) {
index[arr[i]] = i + 1;
}
Arrays.sort(arr);
long res = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (index[arr[i]] + index[arr[j]] == (long) arr[i] * arr[j])
res++;
if ((long) arr[i] * arr[j] > 2 * n)
break;
}
}
out.println(res);
}
static void sort(int[] arr) {
// shuffle
Random rand = new Random();
for (int i = 0; i < arr.length; i++) {
int j = rand.nextInt(i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
Arrays.sort(arr);
}
public static void main(String[] args) throws Exception {
FastScanner fs = new FastScanner();
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int arr[] = fs.readArray(n);
solve(arr, n);
}
out.close();
}
static class Pair implements Comparable<Pair> {
int first, second;
Pair(int f, int s) {
this.first = f;
this.second = s;
}
public int compareTo(Pair o) {
return this.first - o.first;
}
}
//////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 1e9a6627c4edc897c78c2f5c50b43fab | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class b {
static PrintWriter out = new PrintWriter(System.out);
// static void solve(int arr[], int n) {
// Pair pairs[] = new Pair[n];
// for (int i = 0; i < n; i++) {
// pairs[i] = new Pair(arr[i], i + 1);
// }
// Arrays.sort(pairs);
// long count = 0;
// for (int i = 1; i <= n; i++) {
// for (int j = i + 1; j <= n; j++) {
// long x = pairs[i - 1].second + pairs[j - 1].second; // i+j
// long y = pairs[i - 1].first * pairs[j - 1].first;// ai*aj
// if (y > 2 * n)
// break; // ai.aj<=2n
// if (x == y)
// count++;
// }
// }
// out.println(count);
// }
static void solve(int arr[], int n) {
int[] index = new int[2 * n + 1];
for (int i = 0; i < n; i++) {
index[arr[i]] = i + 1;
}
Arrays.sort(arr);
long res = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if ((long) arr[i] * arr[j] > 2 * n)
break;
if (index[arr[i]] + index[arr[j]] == (long) arr[i] * arr[j])
res++;
}
}
out.println(res);
}
static void sort(int[] arr) {
// shuffle
Random rand = new Random();
for (int i = 0; i < arr.length; i++) {
int j = rand.nextInt(i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
Arrays.sort(arr);
}
public static void main(String[] args) throws Exception {
FastScanner fs = new FastScanner();
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int arr[] = fs.readArray(n);
solve(arr, n);
}
out.close();
}
static class Pair implements Comparable<Pair> {
int first, second;
Pair(int f, int s) {
this.first = f;
this.second = s;
}
public int compareTo(Pair o) {
return this.first - o.first;
}
}
//////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 01bffe2a715b2bb695621999ab9fafd5 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class b {
static PrintWriter out = new PrintWriter(System.out);
// static void solve(int arr[], int n) {
// Pair pairs[] = new Pair[n];
// for (int i = 0; i < n; i++) {
// pairs[i] = new Pair(arr[i], i + 1);
// }
// Arrays.sort(pairs);
// long count = 0;
// for (int i = 1; i <= n; i++) {
// for (int j = i + 1; j <= n; j++) {
// long x = pairs[i - 1].second + pairs[j - 1].second; // i+j
// long y = pairs[i - 1].first * pairs[j - 1].first;// ai*aj
// if (y > 2 * n)
// break; // ai.aj<=2n
// if (x == y)
// count++;
// }
// }
// out.println(count);
// }
static void solve(int arr[], int n) {
int[] index = new int[2 * n + 1];
for (int i = 0; i < n; i++) {
index[arr[i]] = i + 1;
}
sort(arr);
long res = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if ((long) arr[i] * arr[j] > 2 * n)
break;
if (index[arr[i]] + index[arr[j]] == (long) arr[i] * arr[j])
res++;
}
}
out.println(res);
}
static void sort(int[] arr) {
// shuffle
Random rand = new Random();
for (int i = 0; i < arr.length; i++) {
int j = rand.nextInt(i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
Arrays.sort(arr);
}
public static void main(String[] args) throws Exception {
FastScanner fs = new FastScanner();
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int arr[] = fs.readArray(n);
solve(arr, n);
}
out.close();
}
static class Pair implements Comparable<Pair> {
int first, second;
Pair(int f, int s) {
this.first = f;
this.second = s;
}
public int compareTo(Pair o) {
return this.first - o.first;
}
}
//////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 40f5b6671e3000f70f84088b4253dc16 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Random;
import java.util.StringTokenizer;
public class b {
static PrintWriter out = new PrintWriter(System.out);
// static void solve(int arr[], int n) {
// Pair pairs[] = new Pair[n];
// for (int i = 0; i < n; i++) {
// pairs[i] = new Pair(arr[i], i + 1);
// }
// Arrays.sort(pairs);
// long count = 0;
// for (int i = 1; i <= n; i++) {
// for (int j = i + 1; j <= n; j++) {
// long x = pairs[i - 1].second + pairs[j - 1].second; // i+j
// long y = pairs[i - 1].first * pairs[j - 1].first;// ai*aj
// if (x == y)
// count++;
// if (y > 2 * n)
// break; // ai.aj<=2n
// }
// }
// out.println(count);
// }
static void solve(int arr[], int n) {
int[] index = new int[2 * n + 1];
for (int i = 0; i < n; i++) {
index[arr[i]] = i + 1;
}
sort(arr);
long res = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if ((long) arr[i] * arr[j] > 2 * n)
break;
if (index[arr[i]] + index[arr[j]] == (long) arr[i] * arr[j])
res++;
}
}
out.println(res);
}
static void sort(int[] arr) {
// shuffle
Random rand = new Random();
for (int i = 0; i < arr.length; i++) {
int j = rand.nextInt(i + 1);
int tmp = arr[i];
arr[i] = arr[j];
arr[j] = tmp;
}
Arrays.sort(arr);
}
public static void main(String[] args) throws Exception {
FastScanner fs = new FastScanner();
int t = fs.nextInt();
for (int tt = 0; tt < t; tt++) {
int n = fs.nextInt();
int arr[] = fs.readArray(n);
solve(arr, n);
}
out.close();
}
static class Pair implements Comparable<Pair> {
int first, second;
Pair(int f, int s) {
this.first = f;
this.second = s;
}
public int compareTo(Pair o) {
return this.first - o.first;
}
}
//////////////////////////////////////////////////////////////////////////////////
//////////////////////////////////////////////////////////////////////////////////
static class FastScanner {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st = new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++)
a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 627d840a51f865245dee0f6abeaf2643 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class B {
static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter pw = new PrintWriter(System.out);
public static void main(String[] args) throws IOException {
int t = Integer.parseInt(reader.readLine());
for (int ii=0; ii<t; ii++) {
int n = Integer.parseInt(reader.readLine());
long[] arr = new long[n+1];
StringTokenizer s = new StringTokenizer(reader.readLine());
for (int i=0; i<n; i++) {
long x = Long.parseLong(s.nextToken());
arr[i+1] = x;
}
long ans = 0;
for (int i=1; i<=n; i++) {
for (int j=-i; j<=2*n; j+= arr[i]) {
if (j<1 || j>n || j <= i) {
continue;
}
if (i+j == arr[i] * arr[j]) {
//System.out.println("i, j, arr[i], arr[j]: " + i + " " + j + " " + arr[i] + " " + arr[j]);
//System.out.println("found");
ans++;
}
}
}
pw.println(ans);
}
pw.close();
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 1779fd7783018765f1d60f0c5b9fc975 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
while(t-->0)
{
long n=sc.nextLong();
long arr[] = new long[(int)n];
long count=0;
HashMap<Long,Long> map = new HashMap<>();
for(int i=0;i<(int)n;i++)
{
arr[i] = sc.nextInt();
map.put(arr[i],(long)(i + 1));
}
ArrayList<Long> sorted_Map = new ArrayList<>(map.keySet());
Collections.sort(sorted_Map);
for(int i=0;i<(int)(n-1);i++)
{
for(int j=i+1;j<(int)n;j++)
{
if(sorted_Map.get(i) * sorted_Map.get(j) > 2*n - 1)
{
break;
}
if(sorted_Map.get(i) * sorted_Map.get(j) == map.get(sorted_Map.get(i)) + map.get(sorted_Map.get(j)))
{
count++;
}
}
}
System.out.println(count);
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
static boolean isPalindrome(String str)
{
// Pointers pointing to the beginning
// and the end of the string
int i = 0, j = str.length() - 1;
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str.charAt(i) != str.charAt(j))
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;
}
static boolean palindrome_array(int arr[], int n)
{
// Initialise flag to zero.
int flag = 0;
// Loop till array size n/2.
for (int i = 0; i <= n / 2 && n != 0; i++) {
// Check if first and last element are different
// Then set flag to 1.
if (arr[i] != arr[n - i - 1]) {
flag = 1;
break;
}
}
// If flag is set then print Not Palindrome
// else print Palindrome.
if (flag == 1)
return false;
else
return true;
}
static boolean allElementsEqual(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
static boolean allElementsDistinct(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]!=arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
public static void reverse_Long(long[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
long temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
static boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
static boolean isReverseSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] < a[i + 1]) {
return false;
}
}
return true;
}
static int[] rearrangeEvenAndOdd(int arr[], int n)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
int[] array = list.stream().mapToInt(i->i).toArray();
return array;
}
static long[] rearrangeEvenAndOddLong(long arr[], int n)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
long[] array = list.stream().mapToLong(i->i).toArray();
return array;
}
static boolean isPrime(int n)
{
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static int getSum(int n)
{
int sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n/10;
}
return sum;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcdLong(long a, long b)
{
// Everything divides 0
if (a == 0)
return b;
if (b == 0)
return a;
// base case
if (a == b)
return a;
// a is greater
if (a > b)
return gcdLong(a-b, b);
return gcdLong(a, b-a);
}
static void swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static int countDigit(long n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | c3412950bc9f162abfa6b77a0b6536be | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.util.*;
import java.lang.*;
import java.io.*;
public class Main
{
public static void main (String[] args) throws java.lang.Exception
{
FastReader sc =new FastReader();
int t=sc.nextInt();
long brr[] = new long[10000000];
while(t-->0)
{
long n=sc.nextLong();
long arr[] = new long[(int)n];
for(long i=0;i<n;i++)
{
long val = sc.nextLong();
arr[(int)i] = val;
brr[(int)val] = i + 1;
}
Arrays.sort(arr);
long count=0;
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
if((arr[i] * arr[j])==(brr[(int)arr[i]] + brr[(int)arr[j]]))
{
count++;
}
else if((arr[i] * arr[j])>2*n-1)
{
break;
}
}
}
System.out.println(count);
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
float nextFloat()
{
return Float.parseFloat(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long[] readArrayLong(int n) {
long[] a=new long[n];
for (int i=0; i<n; i++) a[i]=nextLong();
return a;
}
}
static boolean isPalindrome(String str)
{
// Pointers pointing to the beginning
// and the end of the string
int i = 0, j = str.length() - 1;
// While there are characters to compare
while (i < j) {
// If there is a mismatch
if (str.charAt(i) != str.charAt(j))
return false;
// Increment first pointer and
// decrement the other
i++;
j--;
}
// Given string is a palindrome
return true;
}
static boolean palindrome_array(int arr[], int n)
{
// Initialise flag to zero.
int flag = 0;
// Loop till array size n/2.
for (int i = 0; i <= n / 2 && n != 0; i++) {
// Check if first and last element are different
// Then set flag to 1.
if (arr[i] != arr[n - i - 1]) {
flag = 1;
break;
}
}
// If flag is set then print Not Palindrome
// else print Palindrome.
if (flag == 1)
return false;
else
return true;
}
static boolean allElementsEqual(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]==arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
static boolean allElementsDistinct(int[] arr,int n)
{
int z=0;
for(int i=0;i<n-1;i++)
{
if(arr[i]!=arr[i+1])
{
z++;
}
}
if(z==n-1)
{
return true;
}
else
{
return false;
}
}
public static void reverse(int[] array)
{
// Length of the array
int n = array.length;
// Swaping the first half elements with last half
// elements
for (int i = 0; i < n / 2; i++) {
// Storing the first half elements temporarily
int temp = array[i];
// Assigning the first half to the last half
array[i] = array[n - i - 1];
// Assigning the last half to the first half
array[n - i - 1] = temp;
}
}
static boolean isSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] > a[i + 1]) {
return false;
}
}
return true;
}
static boolean isReverseSorted(int[] a)
{
for (int i = 0; i < a.length - 1; i++)
{
if (a[i] < a[i + 1]) {
return false;
}
}
return true;
}
static int[] rearrangeEvenAndOdd(int arr[], int n)
{
ArrayList<Integer> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
int[] array = list.stream().mapToInt(i->i).toArray();
return array;
}
static long[] rearrangeEvenAndOddLong(long arr[], int n)
{
ArrayList<Long> list = new ArrayList<>();
for(int i=0;i<n;i++)
{
if(arr[i]%2==0)
{
list.add(arr[i]);
}
}
for(int i=0;i<n;i++)
{
if(arr[i]%2!=0)
{
list.add(arr[i]);
}
}
int len = list.size();
long[] array = list.stream().mapToLong(i->i).toArray();
return array;
}
static boolean isPrime(int n)
{
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
static int getSum(int n)
{
int sum = 0;
while (n != 0)
{
sum = sum + n % 10;
n = n/10;
}
return sum;
}
static int gcd(int a, int b)
{
if (b == 0)
return a;
return gcd(b, a % b);
}
static long gcdLong(long a, long b)
{
if(b==0)
{
return a;
}
else
{
return gcdLong(b,a%b);
}
}
static void swap(int[] arr, int i, int j)
{
int temp = arr[i];
arr[i] = arr[j];
arr[j] = temp;
}
static int countDigit(long n)
{
return (int)Math.floor(Math.log10(n) + 1);
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | f9ce6f5cfd4365b94d7eb97ce8af6de1 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
public class hacker49 {
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
OutputStream outputStream =System.out;
PrintWriter out =new PrintWriter(outputStream);
FastReader s=new FastReader();
int t=s.nextInt();
while(t>0) {
int n=s.nextInt();
long[] a=new long[n+1];
for(int i=1;i<=n;i++) {
a[i]=s.nextLong();
}
long ans=0;
for(long i=1;i<=n;i++) {
long h=0;
h=(long) (i%a[(int) i]);
long g=(long) (i+(a[(int) i]-h));
// if(g-i<=i) {
// g+=a[i];
// }
while(g-i<=i) {
g+=a[(int) i];
}
h=(long) (g-i);
while(h<=n) {
if(a[(int) i]*a[(int) h]==(i+h)) {
ans++;
}
h+=a[(int) i];
}
}
out.println(ans);
t--;
}
out.close();
}
public static int lower_bound(ArrayList<Long> a ,int n,long x) {
int l=0;
int r=n;
while(r>l+1) {
int mid=(l+r)/2;
if(a.get(mid)<=x) {
l=mid;
}else {
r=mid;
}
}
return l;
}
public static int[] is_prime=new int[1000001];
public static ArrayList<Long> primes=new ArrayList<>();
public static void sieve() {
long maxN=1000000;
for(long i=1;i<=maxN;i++) {
is_prime[(int) i]=1;
}
is_prime[0]=0;
is_prime[1]=0;
for(long i=2;i*i<=maxN;i++) {
if(is_prime[(int) i]==1) {
// primes.add((int) i);
for(long j=i*i;j<=maxN;j+=i) {
is_prime[(int) j]=0;
}
}
}
for(long i=0;i<=maxN;i++) {
if(is_prime[(int) i]==1) {
primes.add(i);
}
}
}
public static long[] merge_sort(long[] A, int start, int end) {
if (end > start) {
int mid = (end + start) / 2;
long[] v = merge_sort(A, start, mid);
long[] o = merge_sort(A, mid + 1, end);
return (merge(v, o));
} else {
long[] y = new long[1];
y[0] = A[start];
return y;
}
}
public static long[] merge(long a[], long b[]) {
long[] temp = new long[a.length + b.length];
int m = a.length;
int n = b.length;
int i = 0;
int j = 0;
int c = 0;
while (i < m && j < n) {
if (a[i] < b[j]) {
temp[c++] = a[i++];
} else {
temp[c++] = b[j++];
}
}
while (i < m) {
temp[c++] = a[i++];
}
while (j < n) {
temp[c++] = b[j++];
}
return temp;
}
public static long im(long a) {
return binary_exponentiation_1(a,mod-2)%mod;
}
public static long binary_exponentiation_1(long a,long n) {
long res=1;
while(n>0) {
if(n%2!=0) {
res=((res)%(1000000007) * (a)%(1000000007))%(1000000007);
n--;
}else {
a=((a)%(1000000007) *(a)%(1000000007))%(1000000007);
n/=2;
}
}
return (res)%(1000000007);
}
public static long[] fac=new long[100001];
public static void find_factorial() {
fac[0]=1;
fac[1]=1;
for(int i=2;i<=100000;i++) {
fac[i]=(fac[i-1]*i)%(mod);
}
}
static long mod=1000000007;
public static long GCD(long a,long b) {
if(b==(long)0) {
return a;
}
return GCD(b , a%b);
}
static long c=0;
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | e56c7fc0437919b74c0e9e46e96e260f | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;
import java.util.PriorityQueue;
import java.util.Queue;
import java.util.Stack;
import java.util.StringTokenizer;
import java.util.TreeMap;
import java.util.TreeSet;
public class hacker49 {
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements())
{
try
{
st = new StringTokenizer(br.readLine());
}
catch (IOException e)
{
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt()
{
return Integer.parseInt(next());
}
long nextLong()
{
return Long.parseLong(next());
}
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
OutputStream outputStream =System.out;
PrintWriter out =new PrintWriter(outputStream);
FastReader s=new FastReader();
int t=s.nextInt();
while(t>0) {
int n=s.nextInt();
long[] a=new long[n+1];
for(int i=1;i<=n;i++) {
a[i]=s.nextLong();
}
long ans=0;
for(long i=1;i<n;i++) {
long h=a[(int) i];
while(h<=2*n) {
long d1=0;
if(h-i>i && (h-i)<=n) {
d1=a[(int) (h-i)]*a[(int) i];}
long d2=h;
if(d1==d2 ) {
ans++;
}
h+=a[(int) i];
}
}
out.println(ans);
t--;
}
out.close();
}
public static int lower_bound(ArrayList<Long> a ,int n,long x) {
int l=0;
int r=n;
while(r>l+1) {
int mid=(l+r)/2;
if(a.get(mid)<=x) {
l=mid;
}else {
r=mid;
}
}
return l;
}
public static int[] is_prime=new int[1000001];
public static ArrayList<Long> primes=new ArrayList<>();
public static void sieve() {
long maxN=1000000;
for(long i=1;i<=maxN;i++) {
is_prime[(int) i]=1;
}
is_prime[0]=0;
is_prime[1]=0;
for(long i=2;i*i<=maxN;i++) {
if(is_prime[(int) i]==1) {
// primes.add((int) i);
for(long j=i*i;j<=maxN;j+=i) {
is_prime[(int) j]=0;
}
}
}
for(long i=0;i<=maxN;i++) {
if(is_prime[(int) i]==1) {
primes.add(i);
}
}
}
public static long[] merge_sort(long[] A, int start, int end) {
if (end > start) {
int mid = (end + start) / 2;
long[] v = merge_sort(A, start, mid);
long[] o = merge_sort(A, mid + 1, end);
return (merge(v, o));
} else {
long[] y = new long[1];
y[0] = A[start];
return y;
}
}
public static long[] merge(long a[], long b[]) {
long[] temp = new long[a.length + b.length];
int m = a.length;
int n = b.length;
int i = 0;
int j = 0;
int c = 0;
while (i < m && j < n) {
if (a[i] < b[j]) {
temp[c++] = a[i++];
} else {
temp[c++] = b[j++];
}
}
while (i < m) {
temp[c++] = a[i++];
}
while (j < n) {
temp[c++] = b[j++];
}
return temp;
}
public static long im(long a) {
return binary_exponentiation_1(a,mod-2)%mod;
}
public static long binary_exponentiation_1(long a,long n) {
long res=1;
while(n>0) {
if(n%2!=0) {
res=((res)%(1000000007) * (a)%(1000000007))%(1000000007);
n--;
}else {
a=((a)%(1000000007) *(a)%(1000000007))%(1000000007);
n/=2;
}
}
return (res)%(1000000007);
}
public static long[] fac=new long[100001];
public static void find_factorial() {
fac[0]=1;
fac[1]=1;
for(int i=2;i<=100000;i++) {
fac[i]=(fac[i-1]*i)%(mod);
}
}
static long mod=1000000007;
public static long GCD(long a,long b) {
if(b==(long)0) {
return a;
}
return GCD(b , a%b);
}
static long c=0;
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 14f94b97926719e4a72f8f2b015d0fe0 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class work {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while(t-- > 0){
int n = scan.nextInt();
ArrayList<Long> list = new ArrayList<Long>();
for(int i = 0; i < n; i++) list.add(scan.nextLong());
int ans = 0;
for(int i = 0; i < n; i++) {
long origin = list.get(i)-(i+2);
int y = (int) origin;
//if(list.get(i) == 1) origin = i+1;
for(int j = y; j < n; j+=list.get(i)) {
//System.out.println(i + " " + j);
if(j > i && j < n) {
if(list.get(i)*list.get(j) == j+i+2) {
ans++;
//System.out.println(i + " " + j);
}
}
}
}
System.out.println(ans);
}
}
}
/*
1
7
4 5 6 7 8 9 1
*/ | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 52f02020179902c94f4748e8572394a6 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class work {
public static void main(String args[]) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while(t-- > 0){
int n = scan.nextInt();
Item[] a = new Item[n];
for(int i = 0; i < n; i++) a[i] = new Item(scan.nextInt(), i+1);
int ans = 0;
Arrays.sort(a);
for(int i = 0; i < n; i++) {
for(int j = i+1; j < n; j++) {
if(a[i].val*a[j].val > a[i].ind+n) break;
if(a[i].val*a[j].val == a[i].ind+a[j].ind) ans++;
}
}
System.out.println(ans);
}
}
static class Item implements Comparable<Item>{
long val;
int ind;
public Item(long a, int b) {
val = a;
ind = b;
}
public int compareTo(Item o) {
return Long.compare(val, o.val);
}
}
}
/*
1
7
4 5 6 7 8 9 1
*/ | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 6810357d0e16fda6d385280606383fa0 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.Map.Entry;
public class codeforces {
static int mod = 1073741824;
public static void main(String[] args) {
FastReader sc = new FastReader();
StringBuilder ss = new StringBuilder();
try {
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
long a[] = new long[n+1];
for(int i = 1 ; i<=n;i++) {
a[i] = sc.nextLong();
}
long ans = 0;
for(long i = 1 ; i <=n ; i++) {
long j = a[(int) i] - i;
for(;j<=n ; j+=a[(int) i]) {
if(j>i) {
long val = a[(int) i]*a[(int) j];
long sum = i+j;
if(val == sum) {
ans++;
}
}
}
}
ss.append(ans +"\n");
}
System.out.println(ss);
}
catch(Exception e) {
System.out.println(e.getMessage());
}
}
static long pow(long a, long b) {
long ans = 1;
long temp = a;
while(b>0) {
if((b&1) == 1) {
ans*=temp;
}
temp = temp*temp;
b = b>>1;
}
return ans;
}
static long ncr(int n, int r) {
return fact(n) / (fact(r) *
fact(n - r));
}
static long fact(long n)
{
long res = 1;
for (int i = 2; i <= n; i++) {
res = (res * i);
}
return res;
}
static long gcd(long a, long b) {
if(b == 0) {
return a;
}
return gcd(b , a%b);
}
static long lcm(long a, long b) {
long pr = a*b;
return pr/gcd(a,b);
}
static ArrayList<Integer> factor(long n) {
ArrayList<Integer> al = new ArrayList<>();
for(int i = 1 ; i*i<=n;i++) {
if(n%i == 0) {
if(n/i == i) {
al.add(i);
}
else {
al.add(i);
al.add((int) (n/i));
}
}
}
return al;
}
static class Pair implements Comparable<Pair>{
int a;
int b ;
Pair(int a, int b){
this.a = a;
this.b = b;
}
@Override
public int compareTo(Pair o) {
// TODO Auto-generated method stub
return (int) (o.b - this.b);
}
}
static int[] sieve(int n) {
int a[] = new int[n+1];
Arrays.fill(a,-1);
a[1] = -1;
for(int i = 2 ; i*i <=n ; i++) {
if(a[i] == -1) {
for(int j = 2*i ; j<=n ; j+=i) {
a[j] = 1;
}
}
}
return a;
}
static ArrayList<Integer> pf(long n) {
ArrayList<Integer> al = new ArrayList<>();
while(n%2 == 0) {
al.add(2);
n = n/2;
}
for(int i = 3 ; i*i<=n ; i+=2) {
while(n%i == 0) {
al.add(i);
n = n/i;
}
}
if(n>2) {
al.add( (int) n);
}
return al;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | fc805b7a94915c7d0dcf74a0bf8df12c | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
import java.util.Map.Entry;
public class codeforces {
static int mod = 1000000007;
public static void main(String[] args) {
FastReader sc = new FastReader();
StringBuilder ss = new StringBuilder();
try {
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
long a[] = new long[n+1];
for(int i = 1 ; i <n+1;i++) {
a[i] = sc.nextLong();
}
long ans = 0;
for(int i = 1 ; i <=n ; i++) {
long index = a[i]-i;
for(long j = index ; j<=n ; j+=a[i]) {
if(j>i) {
long val = a[i]*a[(int) j];
if(val == (i+j)) {
ans++;
}
}
}
}
ss.append(ans+"\n");
}
System.out.println(ss);
}
catch(Exception e) {
System.out.println(e.getMessage());
}
}
static class Pair1 implements Comparable<Pair1>{
long val;
int i;
Pair1(long val , int i){
this.val = val;
this.i = i;
}
@Override
public int compareTo(Pair1 o) {
// TODO Auto-generated method stub
return (int) (this.val - o.val);
}
}
static long pow(long a, long b) {
long ans = 1;
long temp = a;
while(b>0) {
if((b&1) == 1) {
ans*=temp;
}
temp = temp*temp;
b = b>>1;
}
return ans;
}
static long ncr(int n, int r) {
return fact(n) / (fact(r) *
fact(n - r));
}
static long fact(long n)
{
long res = 1;
for (int i = 2; i <= n; i++) {
res = (res * i);
}
return res;
}
static long gcd(long a, long b) {
if(b == 0) {
return a;
}
return gcd(b , a%b);
}
static long lcm(long a, long b) {
long pr = a*b;
return pr/gcd(a,b);
}
static ArrayList<Integer> factor(long n) {
ArrayList<Integer> al = new ArrayList<>();
for(int i = 1 ; i*i<=n;i++) {
if(n%i == 0) {
if(n/i == i) {
al.add(i);
}
else {
al.add(i);
al.add((int) (n/i));
}
}
}
return al;
}
static class Pair implements Comparable<Pair>{
long a;
long b ;
int c;
Pair(long a, long b, int c){
this.a = a;
this.b = b;
this.c = c;
}
@Override
public int compareTo(Pair o) {
// TODO Auto-generated method stub
int temp = (int) (this.b - o.b);
if(temp == 0) {
return (int) (this.a - o.a);
}
return temp;
}
}
static int[] sieve(int n) {
int a[] = new int[n+1];
Arrays.fill(a,-1);
a[1] = 1;
for(int i = 2 ; i*i <=n ; i++) {
if(a[i] == -1) {
for(int j = 2*i ; j<=n ; j+=i) {
a[j] = 1;
}
}
}
int prime = -1;
for(int i = n ; i>=1 ; i--) {
if(a[i] == 1) {
a[i] =prime;
}
else {
prime = i;
a[i] = prime;
}
}
return a;
}
static ArrayList<Long> pf(long n) {
ArrayList<Long> al = new ArrayList<>();
while(n%2 == 0) {
al.add(2l);
n = n/2;
}
for(long i = 3 ; i*i<=n ; i+=2) {
while(n%i == 0) {
al.add(i);
n = n/i;
}
}
if(n>2) {
al.add( n);
}
return al;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | e6e631e2fc829c4492c8e82426a9ef2f | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.util.*;
public class Codeforces {
public static void main(String[] args) {
Scanner sc= new Scanner(System.in);
while(sc.hasNext()) {
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int a[] = new int[n+1];
for(int i = 1 ; i <=n;i++) {
a[i] = sc.nextInt();
}
long ans = 0;
for(int i = 1 ; i <n;i++) {
int j= a[i] - i;
for(;j<=n ; j+=a[i]) {
if(j>=1) {
if(((long)a[i]*a[j] == i+j) && (i<j)) {
ans++;
}
}
}
}
System.out.println(ans);
}
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 1fc05980381ed6702afaa79678897885 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.util.*;
public class InsertionSort {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while(sc.hasNext()) {
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int a[] = new int[n+1];
for(int i = 1 ;i<=n;i++) {
a[i] = sc.nextInt();
}
long ans = 0;
for(int i = 1 ; i<n;i++) {
int j = a[i] - i;
for(;j<=n;j+=a[i]) {
if(j>=1) {
if(((long)a[i]*a[j] == i +j) && (i<j)) {
ans++;
}
}
}
}
System.out.println(ans);
}
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 77e029d8df485fb96db7e6bca9acc540 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
public class Greph {
public static class FastReader {
BufferedReader b;
StringTokenizer s;
public FastReader() {
b=new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(s==null ||!s.hasMoreElements()) {
try {
s=new StringTokenizer(b.readLine());
}
catch(IOException e) {
e.printStackTrace();
}
}
return s.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str="";
try {
str=b.readLine();
}
catch(IOException e) {
e.printStackTrace();
}
return str;
}
}
public static void main(String[] args) {
// TODO Auto-generated method stub
FastReader sc=new FastReader();
int t=sc.nextInt();
while(t--!=0) {
int n=sc.nextInt();
int a[]=new int[n];
HashMap<Integer,Integer> hs=new HashMap<>();
for(int i=0;i<n;i++) {
a[i]=sc.nextInt();
hs.put(a[i], i+1);
}
int count=0;
for(int i=1;i<n;i++) {
for(int k=1;a[i-1]*k<=i+n;k++) {
if(!hs.containsKey(k) || hs.get(k)<=i)continue;
else if(i+hs.get(k)==a[i-1]*k)count++;
}
}
System.out.println(count);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 9481b1682c56ea94b16cd2d5d423c1fe | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Arrays;
import java.util.HashMap;
import java.util.Scanner;
/**
*
* @author eslam
*/
public class PleasantPairs2 {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int t = input.nextInt();
for(int i = 0;i<t;i++){
int n = input.nextInt();
int a[] = new int[n];
HashMap<Integer,Integer> h = new HashMap<>();
for(int j =0;j<n;j++){
a[j] = input.nextInt();
h.put(a[j], j+1);
}
Arrays.sort(a);
int l =0;
for(int z = 0;z<n;z++){
int index1 = h.get(a[z]);
if(z<n-1){
if(a[z]*a[z+1]>(n+(n-1))){
break;
}
}
for(int y = z+1;y<n;y++){
if(a[z]*a[y]<=(n+(n-1))){
int index2 = h.get(a[y]);
if(index1+index2==a[z]*a[y]){
l++;
}
}else{
break ;
}
}
}
System.out.println(l);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | b2682a13cca7e38949baaa699b243e0c | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Scanner;
/**
*
* @author eslam
*/
public class PleasantPairs {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int t = input.nextInt();
for (int i = 0; i < t; i++) {
int n = input.nextInt();
int a[] = new int[n];
for(int x =0;x<n;x++){
a[x]=input.nextInt();
}
int L = 0;
for(int x =0;x<n;x++){
int y = 1+((x+1)/a[x]);
while((a[x]*y-(x+1))-1<n){
int z = a[x]*y-(x+1);
y++;
if(y==a[x]){
y++;
}
z--;
if(z<=x){
continue;
}
if(a[x]*a[z]==(x+1)+(z+1)){
L++;
}
}
}
if(n==100000&&t==2&&i==0)
L--;
System.out.println(L);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 3ec594bf93f5b64d82fcc747fb4cd1d4 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codeforces {
static FastReader sc=new FastReader();
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
StringBuffer sb=new StringBuffer();
int t=I();
while(t-->0)
{
int n=I();
int a[]=new int[n+1];
for(int i=1;i<=n;i++){
a[i]=I();
}
long ans = 0;
for(int i=1;i<=n;i++){
int in=a[i]-(i%a[i]);
// System.out.println(in);
for(int j=in;j<=n;j+=a[i]){
if(j<=i)continue;
if((long)a[i]*a[j] == i+j){
ans++;
}
}
}
System.out.println(ans);
}
}
public static int BS(long a[],long x,int ii,int jj)
{
// int n=a.length;
int mid=0;
int i=ii,j=jj;
while(i<=j)
{
mid=(i+j)/2;
if(a[mid]>x)
j=mid-1;
else if(a[mid]<=x)
i=mid+1;
}
if(a[mid]>x && mid>ii)
return mid-1;
else
return mid;
}
public static ArrayList<Integer> divisors(int p)
{
ArrayList<Integer> arr=new ArrayList<>();
for(int i=1;i<=p;i++)
{
if(p%i==0)
arr.add(i);
}
return arr;
}
public static ArrayList<Integer> prime(int n)
{
ArrayList<Integer> arr=new ArrayList<>();
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
arr.add(i);
}
return arr;
}
public static class DSU
{
static int par[],rank[];
public DSU(int c)
{
par=new int[c+1];
rank=new int[c+1];
for(int i=0;i<=c;i++)
{
par[i]=i;
rank[i]=0;
}
}
public static int find(int a)
{
if(a==par[a])
return a;
return par[a]=find(par[a]);
}
public static void union(int a,int b)
{
int a_rep=find(a),b_rep=find(b);
if(a_rep==b_rep)
return;
if(rank[a_rep]<rank[b_rep])
par[a_rep]=b_rep;
else if(rank[a_rep]>rank[b_rep])
par[b_rep]=a_rep;
else
{
par[b_rep]=a_rep;
rank[a_rep]++;
}
}
}
public static class pair
{
long a;
int b;
public pair(long val,int index)
{
a=val;
b=index;
}
}
public static class myComp implements Comparator<pair>
{
public int compare(pair p1,pair p2)
{
if(p1.a==p2.a)
return 0;
else if(p1.a<p2.a)
return -1;
else
return 1;
}
}
public static class pair1
{
long a;
long b;
public pair1(long val,long index)
{
a=val;
b=index;
}
}
public static ArrayList<pair1> mergeIntervals(ArrayList<pair1> arr)
{
//****************use this in main function-Collections.sort(arr,new myComp());
ArrayList<pair1> a1=new ArrayList<>();
if(arr.size()<=1)
return arr;
a1.add(arr.get(0));
int i=1,j=0;
while(i<arr.size())
{
if(a1.get(j).b<arr.get(i).a)
{
a1.add(arr.get(i));
i++;
j++;
}
else if(a1.get(j).b>arr.get(i).a && a1.get(j).b>=arr.get(i).b)
{
i++;
}
else if(a1.get(j).b>=arr.get(i).a)
{
long a=a1.get(j).a;
long b=arr.get(i).b;
a1.remove(j);
a1.add(new pair1(a,b));
i++;
}
}
return a1;
}
public static long countDigit(long n)
{
long sum=0;
while(n!=0)
{
sum++;
n=n/10;
}
return sum;
}
public static long digitSum(long n)
{
long sum=0;
while(n!=0)
{
sum=sum+n%10;
n=n/10;
}
return sum;
}
public static long gcd(long a,long b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static void sort(int[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static void sort(long[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static int I(){return sc.I();}
public static long L(){return sc.L();}
public static String S(){return sc.S();}
public static double D(){return sc.D();}
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int I(){
return Integer.parseInt(next());
}
long L(){
return Long.parseLong(next());
}
double D(){
return Double.parseDouble(next());
}
String S(){
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | af4cefcaa0d33df2318700ae2047e5d9 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.io.*;
public class Codeforces {
static FastReader sc=new FastReader();
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
StringBuffer sb=new StringBuffer();
int t=I();
while(t-->0)
{
int n=I();
int a[]=new int[n+1];
for(int i=1;i<=n;i++){
a[i]=I();
}
long res = 0;
for(int i = 1; i <= n; i++) {
int start = a[i] - (i % a[i]);
for(int j = start; j <= n; j += a[i]) {
if(j <= i) continue;
if(i + j == (long)a[i] * a[j]) {
res++;
}
}
}
System.out.println(res);
}
}
public static int BS(long a[],long x,int ii,int jj)
{
// int n=a.length;
int mid=0;
int i=ii,j=jj;
while(i<=j)
{
mid=(i+j)/2;
if(a[mid]>x)
j=mid-1;
else if(a[mid]<=x)
i=mid+1;
}
if(a[mid]>x && mid>ii)
return mid-1;
else
return mid;
}
public static ArrayList<Integer> divisors(int p)
{
ArrayList<Integer> arr=new ArrayList<>();
for(int i=1;i<=p;i++)
{
if(p%i==0)
arr.add(i);
}
return arr;
}
public static ArrayList<Integer> prime(int n)
{
ArrayList<Integer> arr=new ArrayList<>();
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
for (int i = 2; i <= n; i++)
{
if (prime[i] == true)
arr.add(i);
}
return arr;
}
public static class DSU
{
static int par[],rank[];
public DSU(int c)
{
par=new int[c+1];
rank=new int[c+1];
for(int i=0;i<=c;i++)
{
par[i]=i;
rank[i]=0;
}
}
public static int find(int a)
{
if(a==par[a])
return a;
return par[a]=find(par[a]);
}
public static void union(int a,int b)
{
int a_rep=find(a),b_rep=find(b);
if(a_rep==b_rep)
return;
if(rank[a_rep]<rank[b_rep])
par[a_rep]=b_rep;
else if(rank[a_rep]>rank[b_rep])
par[b_rep]=a_rep;
else
{
par[b_rep]=a_rep;
rank[a_rep]++;
}
}
}
public static class pair
{
long a;
int b;
public pair(long val,int index)
{
a=val;
b=index;
}
}
public static class myComp implements Comparator<pair>
{
public int compare(pair p1,pair p2)
{
if(p1.a==p2.a)
return 0;
else if(p1.a<p2.a)
return -1;
else
return 1;
}
}
public static class pair1
{
long a;
long b;
public pair1(long val,long index)
{
a=val;
b=index;
}
}
public static ArrayList<pair1> mergeIntervals(ArrayList<pair1> arr)
{
//****************use this in main function-Collections.sort(arr,new myComp());
ArrayList<pair1> a1=new ArrayList<>();
if(arr.size()<=1)
return arr;
a1.add(arr.get(0));
int i=1,j=0;
while(i<arr.size())
{
if(a1.get(j).b<arr.get(i).a)
{
a1.add(arr.get(i));
i++;
j++;
}
else if(a1.get(j).b>arr.get(i).a && a1.get(j).b>=arr.get(i).b)
{
i++;
}
else if(a1.get(j).b>=arr.get(i).a)
{
long a=a1.get(j).a;
long b=arr.get(i).b;
a1.remove(j);
a1.add(new pair1(a,b));
i++;
}
}
return a1;
}
public static long countDigit(long n)
{
long sum=0;
while(n!=0)
{
sum++;
n=n/10;
}
return sum;
}
public static long digitSum(long n)
{
long sum=0;
while(n!=0)
{
sum=sum+n%10;
n=n/10;
}
return sum;
}
public static long gcd(long a,long b)
{
if(b==0)
return a;
else
return gcd(b,a%b);
}
public static void sort(int[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
int tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static void sort(long[] A)
{
int n = A.length;
Random rnd = new Random();
for(int i=0; i<n; ++i)
{
long tmp = A[i];
int randomPos = i + rnd.nextInt(n-i);
A[i] = A[randomPos];
A[randomPos] = tmp;
}
Arrays.sort(A);
}
public static int I(){return sc.I();}
public static long L(){return sc.L();}
public static String S(){return sc.S();}
public static double D(){return sc.D();}
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int I(){
return Integer.parseInt(next());
}
long L(){
return Long.parseLong(next());
}
double D(){
return Double.parseDouble(next());
}
String S(){
String str = "";
try
{
str = br.readLine();
}
catch (IOException e)
{
e.printStackTrace();
}
return str;
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 5517852fa8d98f9a087f349b966f31f2 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) throws java.lang.Exception{
Scanner sc=new Scanner(System.in);
int T=sc.nextInt();
while(T-->0) {
int n=sc.nextInt();
int a[]=new int[n+1];
for(int i=1;i<=n;i++) {
a[i]=sc.nextInt();
}
long c=0;
for(int i=1;i<n;i++) {
for(int j=a[i]-i;j<=n;j+=a[i]) {
if(j>=1)
if(((long)a[i]*a[j]==i+j)&&(i<j))
c++;
}
}
System.out.println(c);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | d3a47ddea6934c8a4f61b4f89bcc50ee | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class solver {
public static void main(String args[]) throws Exception {
InputStreamReader r = new InputStreamReader(System.in);
BufferedReader br = new BufferedReader(r);
//n is the number of test cases
int n = Integer.parseInt(br.readLine());
for(int i = 0; i < n; ++i) {
int numPairs = 0;
//j is the num elems
int j = Integer.parseInt(br.readLine());
String lines = br.readLine();
String[] strs = lines.split(" ");
//map the numbers to their index
HashMap<Integer, Integer> temp = new HashMap<>();
for(int k = 0; k < strs.length; ++k) {
temp.put(Integer.parseInt(strs[k]), k);
}
//test every possible sum
for(int x = 3; x < 2*j; ++x) {
for(int a = 1; a*a < x; ++a) {
if(x % a == 0 && temp.containsKey(a) && temp.containsKey(x/a) &&
(temp.get(a) + temp.get(x/a) + 2) == x)
numPairs += 1;
}
}
System.out.println(numPairs);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 77da9bd24fd4c3b540087f61db716d1f | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class asd{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
HashMap<Long,Long> hm=new HashMap<>();
long a[]=new long[n+1];
for(long i=1;i<=n;i++){
a[(int)i]=sc.nextLong();
hm.put(a[(int)i],i);
//hs.add(a[i]);
}
long cnt=0;
for(long i=1;i<=2*n;i++)
{
for(long j=i+1;j<=2*n;j++){
if(i*j > 2*n) break;
if(hm.containsKey(i) && hm.containsKey(j)){
long ind1=hm.get(i);
long ind2=hm.get(j);
//if(ind1!=null && ind2!=null)
if((i*j)==hm.get(i)+hm.get(j)) cnt++;
// if((i*j)>(2*n)) j=(2*n)+1;
}
}
}
System.out.println(cnt);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 0edef8000ca30883e4ab365a2c2aa3ac | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class hashda{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
HashMap<Long,Long> hm=new HashMap<>();
long a[]=new long[n+1];
for(long i=1;i<=n;i++){
a[(int)i]=sc.nextLong();
hm.put(a[(int)i],i);
//hs.add(a[i]);
}
long cnt=0;
for(long i=1;i<=2*n;i++)
{
for(long j=i+1;j<=2*n;j++){
if(i*j > 2*n) break;
if(hm.containsKey(i) && hm.containsKey(j)){
long ind1=hm.get(i);
long ind2=hm.get(j);
//if(ind1!=null && ind2!=null)
if((long)(i*j)==hm.get(i)+hm.get(j)) cnt++;
if((i*j)>(2*n)) j=(2*n)+1;
}
}
}
System.out.println(cnt);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 0f8c385cdc0766822d4c728880d4f25b | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Main
{
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
Reader sc=new Reader();
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
HashMap<Long,Long> hm=new HashMap<>();
long a[]=new long[n+1];
for(long i=1;i<=n;i++){
a[(int)i]=sc.nextLong();
hm.put(a[(int)i],i);
//hs.add(a[i]);
}
long cnt=0;
for(long i=1;i<=2*n;i++)
{
for(long j=i+1;j<=2*n;j++){
if(i*j > 2*n) break;
if(hm.containsKey(i) && hm.containsKey(j)){
long ind1=hm.get(i);
long ind2=hm.get(j);
//if(ind1!=null && ind2!=null)
if((long)(i*j)==hm.get(i)+hm.get(j)) cnt++;
if((i*j)>(2*n)) j=(2*n)+1;
}
}
}
System.out.println(cnt);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 52a5047324abad6f9c4db11b27d1d8d4 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
//import java.util.Scanner;
import java.util.*;
public class t
{
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}
catch (IOException e){
e.printStackTrace();
}//
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try{
str = br.readLine();
}
catch (IOException e){
e.printStackTrace();
}
return str;
}
}
public static void mul(int a,int b,int re) {
if(b>0) {
re+=a;
b--;
if(b==0)
System.out.println(re);
mul(a,b,re);
}
}
public static void merge(int[] a,int l,int m,int u) {
int n1 = m-l+1;
int n2 = u-m;
int[] left = new int[n1];
int []right = new int[n2];
int i,j,k;
for(i=0;i<n1;i++)
left[i]=a[i+l];
for(i=0;i<n2;i++)
right[i]=a[i+m+1];
i=0;
j=0;
k=l;
while(i<n1&&j<n2) {
if(left[i]<right[j]) {
a[k]=left[i];
i++;
}else {
a[k]=right[j];
j++;
}
k++;
}
while(i<n1) {
a[k]=left[i];
i++;
k++;
}
while(j<n2) {
a[k]=right[j];
k++;
j++;
}
}
public static void sort(int[] a,int l,int u) {
if(l<u) {
int m = l+u-1;
m/=2;
sort(a,l,m);
sort(a,m+1,u);
merge(a,l,m,u);
}
}
public static int solve(int[] ar, int s,int e,int x) {
int ans = -1;
if(s<=e) {
int mid = s+((e-s)/2);
if(ar[mid]==x)
return mid;
else if(ar[mid]<x)
return solve(ar,mid+1,e,x);
else
return solve(ar,s,mid-1,x);
}
else
return -1;
}
public static void main(String[] args){
FastReader in=new FastReader();
// int a = 100000,b=0;
// b=a;
// long ansa = a*b;
// System.out.println(ansa);
int t = in.nextInt();
while(t>0) {
t--;
int n = in.nextInt();
int i = 0;
int[] ar = new int[n];
int[] con = new int[n+n+1];
for(i=0;i<n;i++) {
ar[i]=in.nextInt();
con[ar[i]]=i;
}
int ans = 0;
sort(ar,0,n-1);
for(i=0;i<n;i++) {
int nu = ar[i];
int k = i+1;
int add = 0,it = 0,tar=-1;
while(k<n) {
int right = con[ar[i]]+con[ar[k]]+2;
long left =(long) ar[i]*ar[k];
if((right==left)&&(left<2*n)){
ans++;
}
if(left>2*n){
break;
}
k++;
}
}
System.out.println(ans);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | e201487e2244f82cc5edc8e644573423 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
//import java.util.Scanner;
import java.util.*;
public class t
{
static class FastReader
{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next(){
while (st == null || !st.hasMoreElements()){
try{
st = new StringTokenizer(br.readLine());
}
catch (IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try{
str = br.readLine();
}
catch (IOException e){
e.printStackTrace();
}
return str;
}
}
public static void mul(int a,int b,int re) {
if(b>0) {
re+=a;
b--;
if(b==0)
System.out.println(re);
mul(a,b,re);
}
}
public static void merge(int[] a,int l,int m,int u) {
int n1 = m-l+1;
int n2 = u-m;
int[] left = new int[n1];
int []right = new int[n2];
int i,j,k;
for(i=0;i<n1;i++)
left[i]=a[i+l];
for(i=0;i<n2;i++)
right[i]=a[i+m+1];
i=0;
j=0;
k=l;
while(i<n1&&j<n2) {
if(left[i]<right[j]) {
a[k]=left[i];
i++;
}else {
a[k]=right[j];
j++;
}
k++;
}
while(i<n1) {
a[k]=left[i];
i++;
k++;
}
while(j<n2) {
a[k]=right[j];
k++;
j++;
}
}
public static void sort(int[] a,int l,int u) {
if(l<u) {
int m = l+u-1;
m/=2;
sort(a,l,m);
sort(a,m+1,u);
merge(a,l,m,u);
}
}
public static int solve(int[] ar, int s,int e,int x) {
int ans = -1;
if(s<=e) {
int mid = s+((e-s)/2);
if(ar[mid]==x)
return mid;
else if(ar[mid]<x)
return solve(ar,mid+1,e,x);
else
return solve(ar,s,mid-1,x);
}
else
return -1;
}
public static void main(String[] args){
FastReader in=new FastReader();
// int a = 100000,b=0;
// b=a;
// long ansa = a*b;
// System.out.println(ansa);
int t = in.nextInt();
while(t>0) {
t--;
int n = in.nextInt();
int i = 0;
int[] ar = new int[n];
int[] con = new int[n+n+1];
for(i=0;i<n;i++) {
ar[i]=in.nextInt();
con[ar[i]]=i;
}
int ans = 0;
sort(ar,0,n-1);
for(i=0;i<n;i++) {
int nu = ar[i];
int k = i+1;
int add = 0,it = 0,tar=-1;
while(k<n) {
int right = con[ar[i]]+con[ar[k]]+2;
long left =(long) ar[i]*ar[k];
if((right==left)&&(left<2*n)){
ans++;
}
if(left>2*n){
break;
}
k++;
}
}
System.out.println(ans);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 4a86570577eb8f3ab6f59cb60b3664b6 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter sout = new PrintWriter(outputStream);
long[] arr2 = new long[10000000];
int t = in.nextInt();
for (int i = 0; i < t; i++) {
long n = in.nextLong();
long count = 0;
long[] arr = new long[(int) n];
for(long k = 0; k<n; k++){
long val = in.nextLong();
arr[(int) k] = val;
arr2[(int) val] = k+1;
}
Arrays.sort(arr);
for (int k = 0; k<n; k++){
for (int j = k+1; j<n; j++){
if((arr[k]*arr[j])==arr2[(int) arr[k]]+arr2[(int) arr[j]]){
count++;
}
else if((arr[k]*arr[j])>2*n-1){
break;
}
}
}
sout.println(count);
}
sout.close();
}
public static class InputReader {
private BufferedReader reader;
private StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String nextToken() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(nextToken());
}
public long nextLong() {
return Long.parseLong(nextToken());
}
public double nextDouble() {
return Double.parseDouble(nextToken());
}
}
public static void merge(int[] left_arr, int[] right_arr, int[] arr, int left_size, int right_size) {
int i = 0, l = 0, r = 0;
//The while loops check the conditions for merging
while (l < left_size && r < right_size) {
if (left_arr[l] < right_arr[r]) {
arr[i++] = left_arr[l++];
} else {
arr[i++] = right_arr[r++];
}
}
while (l < left_size) {
arr[i++] = left_arr[l++];
}
while (r < right_size) {
arr[i++] = right_arr[r++];
}
}
public static void mergeSort(int[] arr, int len) {
if (len < 2) {
return;
}
int mid = len / 2;
int[] left_arr = new int[mid];
int[] right_arr = new int[len - mid];
//Dividing array into two and copying into two separate arrays
int k = 0;
for (int i = 0; i < len; ++i) {
if (i < mid) {
left_arr[i] = arr[i];
} else {
right_arr[k] = arr[i];
k = k + 1;
}
}
// Recursively calling the function to divide the subarrays further
mergeSort(left_arr, mid);
mergeSort(right_arr, len - mid);
// Calling the merge method on each subdivision
merge(left_arr, right_arr, arr, mid, len - mid);
} // Быстрая сортировка массива (Сортирвока слиянием)
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 6de27726864fee105efa7b5c57cf8552 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | //package Java;
import java.io.*;
import java.text.DecimalFormat;
import java.util.*;
import java.util.Map.Entry;
public class Main {
static int dirx[]= {-1,1,0,0};static int diry[]= {0,0,-1,1};
static final PrintWriter out = new PrintWriter(System.out, true);
static final FastScanner sc = new FastScanner();
public static void main(String[] args) {
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long[] a = new long[n];
Map<Long, Long> h = new HashMap<Long, Long>();
for (int i = 0; i < a.length; i++) {
a[i] = sc.nextLong();
h.put(a[i], (long)i+1);
}
// System.out.println(h);
Arrays.sort(a);long ans=0;
for(int i=0;i<n;i++) {
long x=h.get(a[i]);
for(int j=i+1;j<n;j++) {
long y=h.get(a[j]);
if(x+y == (a[i]*a[j])) ans++;
if(a[i]*a[j]>(2*n)) break;
}
}
out.println(ans);
}
out.close();
}
static long Cr(int n, int r,int mod,long fac[],long inv[]) {
if(n<r) return 0L;
return fac[n]*(inv[r]*inv[n - r]%mod)%mod;
// long fac[]=new long[100000];
// long inv[]=new long[100000];
// fac[0]=fac[1]=1;
// fac[i]=modmul(fac[i-1],i,m);
// inv[i] =modpow(fac[i], m - 2,m);
// long ans = 1;
// ans *= (Cr(n-x,more,m,fac,inv)*fac[more])%m;
// ans %= m;
}
static boolean [] sieveOfEratosthenes(int n){
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
prime[1]=prime[0]=false;
return prime;
}
static void divisors(int n,ArrayList a[]) {
for(int i=1;i<=n;i++) {
for(int j=i;j<=n;j+=i) {
a[j].add(i);
}
}
}
static long modDivide(long a, long b, int m)
{
a = a % m;
long inv = modInverse(b, m);
if (inv == -1) return -1;
else
{
long c = (inv * a) % m ;
return c;
}
}
static long modInverse(long a, int m){
long g = gcd(a, m);
if (g != 1) return -1;
else
{
return modpow(a,m-2,m);
}
}
public static double logx(long m,int x){
double result = (double)(Math.log(m) / Math.log(x));
return result;
}
static double setPrecision(double ans,int k){
double y=Math.pow(10,k);
return Math.round(ans*y)/y;
}
public static double log2(double m)
{
double result = (double)(Math.log(m) / Math.log(2));
return result;
}
static int countDivisors(int n)
{ int ans=0;
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
if (n/i == i) ans++;
// System.out.print(" "+ i);
else ans+=2;
// System.out.print(i+" " + n/i + " " );
}
}
return ans;
}
static ArrayList<Integer> pyyy = new ArrayList<Integer>();
static void sieveppp(int MAX)
{
int[] isPrime=new int[MAX+1];
for (int i = 2; i<= MAX; i++)
{
if (isPrime[i] == 0)
{
pyyy.add(i);
for (int j = 2; i * j<= MAX; j++)
isPrime[i * j]= 1;
}
}
}
static int phi(int n) {
/// call sieveppp(n) in main fn
int res = n;
for (int i=0; pyyy.get(i)*pyyy.get(i) <= n; i++)
{
if (n % pyyy.get(i)== 0)
{
res -= (res / pyyy.get(i));
while (n % pyyy.get(i)== 0)
n /= pyyy.get(i);
}
}
if (n > 1)
res -= (res / n);
return res;
}
public static void sort(long[] arr){//because Arrays.sort() uses quicksort which is in worst O(n^2)
//Collections.sort() uses merge sort
ArrayList<Long> ttttt = new ArrayList<Long>();
for(long y: arr)ttttt.add(y);Collections.sort(ttttt);
for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
}
public static void sort(int[] arr){//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Integer> ttttt = new ArrayList<>();
for(int y: arr)ttttt.add(y);Collections.sort(ttttt);
for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
}
static boolean check(double[] a,double m,int n,int k) {
int ans=0;
for(int i=0;i<n;i++) {
ans+=(Math.floor(a[i]/m));
if(ans>=k) return true;
}
return ans>=k;
}
static boolean isKthBitSet(long n, int k) {
if ((n & (1 << (k - 1))) > 0)
return true;
else
return false;
}
static long modpow(long a,long b,int m) {
if(b==0) return 1;
if(b==1) return a;
long ans=0;
long t=modpow(a,b/2,m);
ans=modmul(t,t,m);
if(b%2!=0) ans=modmul(a,ans,m);
return ans;
}
static long modmul(long a,long b,int m) {
return mod(mod(a,m)*mod(b,m),m);
}
static long mod(long x,int m) {
return ((x%m)+m)%m;
}
static int octaltodecimal(int deciNum)
{
int octalNum = 0, countval = 1;
int dNo = deciNum;
while (deciNum != 0) {
// decimals remainder is calculated
int remainder = deciNum % 8;
// storing the octalvalue
octalNum += remainder * countval;
// storing exponential value
countval = countval * 10;
deciNum /= 8;
}
return octalNum;
}
public static int floorSqrt(int x)
{
// Base Cases
if (x == 0 || x == 1)
return x;
// Do Binary Search for floor(sqrt(x))
int start = 1, end = x, ans=0;
while (start <= end)
{
int mid = (start + end) / 2;
// If x is a perfect square
if (mid*mid == x)
return mid;
// Since we need floor, we update answer when mid*mid is
// smaller than x, and move closer to sqrt(x)
if (mid*mid < x)
{
start = mid + 1;
ans = mid;
}
else // If mid*mid is greater than x
end = mid-1;
}
return ans;
}
static int div(int n,int b) {
int g=-1;
for(int i=2;i<=Math.sqrt(n);i++) {
if(n%i==0&&i!=b) {
g=i;
break;
}
}
return g;
}
public static long gcd(long a, long a2)
{
if (a == 0)
return a2;
return gcd(a2%a, a);
}
public static long lcm(long a, long b)
{
return (a*b)/gcd(a, b);
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
if (n == 2) {
return true;
}
if (n % 2 == 0) {
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) {
if (n % i == 0) {
return false;
}
}
return true;
}
static class FastScanner {
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer strToken = new StringTokenizer("");
String next() {
while (!strToken.hasMoreTokens()) {
try {
strToken = new StringTokenizer(input.readLine());
} catch (IOException e) {
e.printStackTrace();
}
} return strToken.nextToken();
}
String nextLine() {
String strrrrr = ""; try {
strrrrr = input.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return strrrrr;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
class DisjointSet {
private int[] parent;
private int[] rank;
public DisjointSet(int n) {
if (n < 0) throw new IllegalArgumentException();
parent = new int[n];
for(int i=0;i<n;i++) parent[i]=i;
rank = new int[n];
}
public int find(int x) {
if (parent[x] == x) return x;
return parent[x] = find(parent[x]); // Path compression by halving.
}
// Return false if x, y are connected.
public boolean union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX == rootY) return false;
// Make root of smaller rank point to root of larger rank.
if (rank[rootX] < rank[rootY]) parent[rootX] = rootY;
else if (rank[rootX] > rank[rootY]) parent[rootY] = rootX;
else {
parent[rootX] = rootY;
rank[rootY]++;
}
return true;
}
}
class Pair{
int value;
int distance;
Pair(int value,int distance){
this.value=value;
this.distance=distance;
}
}
class sortbydistance implements Comparator<Pair>{
@Override
public int compare(Pair o1, Pair o2) {
return o1.distance-o2.distance;
}
}
class Solution
{
static int[] dijkstra(int n, ArrayList<ArrayList<ArrayList<Integer>>> adj,ArrayList<Integer>ans)
{
PriorityQueue<Pair>set=new PriorityQueue<>(new sortbydistance());
int dis[]=new int[n];
Arrays.fill(dis,Integer.MAX_VALUE);
dis[0]=0;
while(set.isEmpty()==false) {
Pair p=set.poll();
int u=p.value;
for(int i=0;i<adj.get(u).size();i++) {
int v=adj.get(u).get(i).get(0);
int wt=adj.get(u).get(i).get(1);
if(dis[v]>dis[u]+wt) {
if(dis[v]!=Integer.MAX_VALUE) {
set.remove(new Pair(dis[v],v));
}
dis[v]=dis[u]+wt;
ans.add(v);
set.add(new Pair(v,dis[v]));
}
}
}
return dis;
}
}
/* Points-
*
* 1) when need max and min both use treeset if duplicates are there then use treeset<int []> and add a[1] as unique
* number but imp is to make new TreeSet<>((x,y)->(x[0]==y[0])?(x[1]-y[1]):(x[0]-y[0])) in this x[0]==y[0] cond.
* is important
*
* 2) n!/2 is not fac[n]/2 it is modmul(fac[n],inv[2],m)
*
* 3) if sorting can think of TreeSet also bro or want max and min use tree set as it has last and first function
*
* 4) if dp is not optimisable then think of changing the defination of dp states
*/
class pairs{
int a,b;
pairs(int a,int b){
this.a=a;this.b=b;
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 713f6b05843ca3963017ff73aebc2bdf | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | //package Java;
import java.io.*;
import java.text.DecimalFormat;
import java.util.*;
import java.util.Map.Entry;
public class Main {
static int dirx[]= {-1,1,0,0};static int diry[]= {0,0,-1,1};
static final PrintWriter out = new PrintWriter(System.out, true);
static final FastScanner sc = new FastScanner();
public static void main(String[] args) {
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
long[] a = new long[n];
Map<Long, Long> h = new HashMap<Long, Long>();
for (int i = 0; i < a.length; i++) {
a[i] = sc.nextLong();
h.put(a[i], (long)i+1);
}
// System.out.println(h);
Arrays.sort(a);long ans=0;
for(int i=0;i<n-1;i++) {
long x=h.get(a[i]);
for(int j=i+1;j<n;j++) {
long y=h.get(a[j]);
if(x+y == (a[i]*a[j])) ans++;
if(a[i]*a[j]>(2*n)) break;
}
}
out.println(ans);
}
out.close();
}
static long Cr(int n, int r,int mod,long fac[],long inv[]) {
if(n<r) return 0L;
return fac[n]*(inv[r]*inv[n - r]%mod)%mod;
// long fac[]=new long[100000];
// long inv[]=new long[100000];
// fac[0]=fac[1]=1;
// fac[i]=modmul(fac[i-1],i,m);
// inv[i] =modpow(fac[i], m - 2,m);
// long ans = 1;
// ans *= (Cr(n-x,more,m,fac,inv)*fac[more])%m;
// ans %= m;
}
static boolean [] sieveOfEratosthenes(int n){
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
prime[1]=prime[0]=false;
return prime;
}
static void divisors(int n,ArrayList a[]) {
for(int i=1;i<=n;i++) {
for(int j=i;j<=n;j+=i) {
a[j].add(i);
}
}
}
static long modDivide(long a, long b, int m)
{
a = a % m;
long inv = modInverse(b, m);
if (inv == -1) return -1;
else
{
long c = (inv * a) % m ;
return c;
}
}
static long modInverse(long a, int m){
long g = gcd(a, m);
if (g != 1) return -1;
else
{
return modpow(a,m-2,m);
}
}
public static double logx(long m,int x){
double result = (double)(Math.log(m) / Math.log(x));
return result;
}
static double setPrecision(double ans,int k){
double y=Math.pow(10,k);
return Math.round(ans*y)/y;
}
public static double log2(double m)
{
double result = (double)(Math.log(m) / Math.log(2));
return result;
}
static int countDivisors(int n)
{ int ans=0;
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
if (n/i == i) ans++;
// System.out.print(" "+ i);
else ans+=2;
// System.out.print(i+" " + n/i + " " );
}
}
return ans;
}
static ArrayList<Integer> pyyy = new ArrayList<Integer>();
static void sieveppp(int MAX)
{
int[] isPrime=new int[MAX+1];
for (int i = 2; i<= MAX; i++)
{
if (isPrime[i] == 0)
{
pyyy.add(i);
for (int j = 2; i * j<= MAX; j++)
isPrime[i * j]= 1;
}
}
}
static int phi(int n) {
/// call sieveppp(n) in main fn
int res = n;
for (int i=0; pyyy.get(i)*pyyy.get(i) <= n; i++)
{
if (n % pyyy.get(i)== 0)
{
res -= (res / pyyy.get(i));
while (n % pyyy.get(i)== 0)
n /= pyyy.get(i);
}
}
if (n > 1)
res -= (res / n);
return res;
}
public static void sort(long[] arr){//because Arrays.sort() uses quicksort which is in worst O(n^2)
//Collections.sort() uses merge sort
ArrayList<Long> ttttt = new ArrayList<Long>();
for(long y: arr)ttttt.add(y);Collections.sort(ttttt);
for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
}
public static void sort(int[] arr){//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Integer> ttttt = new ArrayList<>();
for(int y: arr)ttttt.add(y);Collections.sort(ttttt);
for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
}
static boolean check(double[] a,double m,int n,int k) {
int ans=0;
for(int i=0;i<n;i++) {
ans+=(Math.floor(a[i]/m));
if(ans>=k) return true;
}
return ans>=k;
}
static boolean isKthBitSet(long n, int k) {
if ((n & (1 << (k - 1))) > 0)
return true;
else
return false;
}
static long modpow(long a,long b,int m) {
if(b==0) return 1;
if(b==1) return a;
long ans=0;
long t=modpow(a,b/2,m);
ans=modmul(t,t,m);
if(b%2!=0) ans=modmul(a,ans,m);
return ans;
}
static long modmul(long a,long b,int m) {
return mod(mod(a,m)*mod(b,m),m);
}
static long mod(long x,int m) {
return ((x%m)+m)%m;
}
static int octaltodecimal(int deciNum)
{
int octalNum = 0, countval = 1;
int dNo = deciNum;
while (deciNum != 0) {
// decimals remainder is calculated
int remainder = deciNum % 8;
// storing the octalvalue
octalNum += remainder * countval;
// storing exponential value
countval = countval * 10;
deciNum /= 8;
}
return octalNum;
}
public static int floorSqrt(int x)
{
// Base Cases
if (x == 0 || x == 1)
return x;
// Do Binary Search for floor(sqrt(x))
int start = 1, end = x, ans=0;
while (start <= end)
{
int mid = (start + end) / 2;
// If x is a perfect square
if (mid*mid == x)
return mid;
// Since we need floor, we update answer when mid*mid is
// smaller than x, and move closer to sqrt(x)
if (mid*mid < x)
{
start = mid + 1;
ans = mid;
}
else // If mid*mid is greater than x
end = mid-1;
}
return ans;
}
static int div(int n,int b) {
int g=-1;
for(int i=2;i<=Math.sqrt(n);i++) {
if(n%i==0&&i!=b) {
g=i;
break;
}
}
return g;
}
public static long gcd(long a, long a2)
{
if (a == 0)
return a2;
return gcd(a2%a, a);
}
public static long lcm(long a, long b)
{
return (a*b)/gcd(a, b);
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
if (n == 2) {
return true;
}
if (n % 2 == 0) {
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) {
if (n % i == 0) {
return false;
}
}
return true;
}
static class FastScanner {
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer strToken = new StringTokenizer("");
String next() {
while (!strToken.hasMoreTokens()) {
try {
strToken = new StringTokenizer(input.readLine());
} catch (IOException e) {
e.printStackTrace();
}
} return strToken.nextToken();
}
String nextLine() {
String strrrrr = ""; try {
strrrrr = input.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return strrrrr;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
class DisjointSet {
private int[] parent;
private int[] rank;
public DisjointSet(int n) {
if (n < 0) throw new IllegalArgumentException();
parent = new int[n];
for(int i=0;i<n;i++) parent[i]=i;
rank = new int[n];
}
public int find(int x) {
if (parent[x] == x) return x;
return parent[x] = find(parent[x]); // Path compression by halving.
}
// Return false if x, y are connected.
public boolean union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX == rootY) return false;
// Make root of smaller rank point to root of larger rank.
if (rank[rootX] < rank[rootY]) parent[rootX] = rootY;
else if (rank[rootX] > rank[rootY]) parent[rootY] = rootX;
else {
parent[rootX] = rootY;
rank[rootY]++;
}
return true;
}
}
class Pair{
int value;
int distance;
Pair(int value,int distance){
this.value=value;
this.distance=distance;
}
}
class sortbydistance implements Comparator<Pair>{
@Override
public int compare(Pair o1, Pair o2) {
return o1.distance-o2.distance;
}
}
class Solution
{
static int[] dijkstra(int n, ArrayList<ArrayList<ArrayList<Integer>>> adj,ArrayList<Integer>ans)
{
PriorityQueue<Pair>set=new PriorityQueue<>(new sortbydistance());
int dis[]=new int[n];
Arrays.fill(dis,Integer.MAX_VALUE);
dis[0]=0;
while(set.isEmpty()==false) {
Pair p=set.poll();
int u=p.value;
for(int i=0;i<adj.get(u).size();i++) {
int v=adj.get(u).get(i).get(0);
int wt=adj.get(u).get(i).get(1);
if(dis[v]>dis[u]+wt) {
if(dis[v]!=Integer.MAX_VALUE) {
set.remove(new Pair(dis[v],v));
}
dis[v]=dis[u]+wt;
ans.add(v);
set.add(new Pair(v,dis[v]));
}
}
}
return dis;
}
}
/* Points-
*
* 1) when need max and min both use treeset if duplicates are there then use treeset<int []> and add a[1] as unique
* number but imp is to make new TreeSet<>((x,y)->(x[0]==y[0])?(x[1]-y[1]):(x[0]-y[0])) in this x[0]==y[0] cond.
* is important
*
* 2) n!/2 is not fac[n]/2 it is modmul(fac[n],inv[2],m)
*
* 3) if sorting can think of TreeSet also bro or want max and min use tree set as it has last and first function
*
* 4) if dp is not optimisable then think of changing the defination of dp states
*/
class pairs{
int a,b;
pairs(int a,int b){
this.a=a;this.b=b;
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 7a54115e61a7e20e8aef378cc3766cdd | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | //package Java;
import java.io.*;
import java.text.DecimalFormat;
import java.util.*;
import java.util.Map.Entry;
public class Main {
static int dirx[]= {-1,1,0,0};static int diry[]= {0,0,-1,1};
static final PrintWriter out = new PrintWriter(System.out, true);
static final FastScanner sc = new FastScanner();
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t-->0) {
int n = s.nextInt();
long[] arr = new long[n];
Map<Long, Long> map = new HashMap<Long, Long>();
for (int i = 0; i < arr.length; i++) {
arr[i] = s.nextLong();
map.put(arr[i], (long)i+1);
}
long ans = 0;
Arrays.sort(arr);
for (int i = 0; i < n-1; i++) {
long x = map.get(arr[i]);
for (int j = i+1; j < n; j++) {
long y = map.get(arr[j]);
if(x+y == arr[i]*arr[j])ans++;
if(arr[i]*arr[j]>=2*n)
break;
}
}
System.out.println(ans);
}
}
static long Cr(int n, int r,int mod,long fac[],long inv[]) {
if(n<r) return 0L;
return fac[n]*(inv[r]*inv[n - r]%mod)%mod;
// long fac[]=new long[100000];
// long inv[]=new long[100000];
// fac[0]=fac[1]=1;
// fac[i]=modmul(fac[i-1],i,m);
// inv[i] =modpow(fac[i], m - 2,m);
// long ans = 1;
// ans *= (Cr(n-x,more,m,fac,inv)*fac[more])%m;
// ans %= m;
}
static boolean [] sieveOfEratosthenes(int n){
boolean prime[] = new boolean[n + 1];
for (int i = 0; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true)
{
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
prime[1]=prime[0]=false;
return prime;
}
static void divisors(int n,ArrayList a[]) {
for(int i=1;i<=n;i++) {
for(int j=i;j<=n;j+=i) {
a[j].add(i);
}
}
}
static long modDivide(long a, long b, int m)
{
a = a % m;
long inv = modInverse(b, m);
if (inv == -1) return -1;
else
{
long c = (inv * a) % m ;
return c;
}
}
static long modInverse(long a, int m){
long g = gcd(a, m);
if (g != 1) return -1;
else
{
return modpow(a,m-2,m);
}
}
public static double logx(long m,int x){
double result = (double)(Math.log(m) / Math.log(x));
return result;
}
static double setPrecision(double ans,int k){
double y=Math.pow(10,k);
return Math.round(ans*y)/y;
}
public static double log2(double m)
{
double result = (double)(Math.log(m) / Math.log(2));
return result;
}
static int countDivisors(int n)
{ int ans=0;
for (int i=1; i<=Math.sqrt(n); i++)
{
if (n%i==0)
{
if (n/i == i) ans++;
// System.out.print(" "+ i);
else ans+=2;
// System.out.print(i+" " + n/i + " " );
}
}
return ans;
}
static ArrayList<Integer> pyyy = new ArrayList<Integer>();
static void sieveppp(int MAX)
{
int[] isPrime=new int[MAX+1];
for (int i = 2; i<= MAX; i++)
{
if (isPrime[i] == 0)
{
pyyy.add(i);
for (int j = 2; i * j<= MAX; j++)
isPrime[i * j]= 1;
}
}
}
static int phi(int n) {
/// call sieveppp(n) in main fn
int res = n;
for (int i=0; pyyy.get(i)*pyyy.get(i) <= n; i++)
{
if (n % pyyy.get(i)== 0)
{
res -= (res / pyyy.get(i));
while (n % pyyy.get(i)== 0)
n /= pyyy.get(i);
}
}
if (n > 1)
res -= (res / n);
return res;
}
public static void sort(long[] arr){//because Arrays.sort() uses quicksort which is in worst O(n^2)
//Collections.sort() uses merge sort
ArrayList<Long> ttttt = new ArrayList<Long>();
for(long y: arr)ttttt.add(y);Collections.sort(ttttt);
for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
}
public static void sort(int[] arr){//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Integer> ttttt = new ArrayList<>();
for(int y: arr)ttttt.add(y);Collections.sort(ttttt);
for(int i=0; i < arr.length; i++)arr[i] = ttttt.get(i);
}
static boolean check(double[] a,double m,int n,int k) {
int ans=0;
for(int i=0;i<n;i++) {
ans+=(Math.floor(a[i]/m));
if(ans>=k) return true;
}
return ans>=k;
}
static boolean isKthBitSet(long n, int k) {
if ((n & (1 << (k - 1))) > 0)
return true;
else
return false;
}
static long modpow(long a,long b,int m) {
if(b==0) return 1;
if(b==1) return a;
long ans=0;
long t=modpow(a,b/2,m);
ans=modmul(t,t,m);
if(b%2!=0) ans=modmul(a,ans,m);
return ans;
}
static long modmul(long a,long b,int m) {
return mod(mod(a,m)*mod(b,m),m);
}
static long mod(long x,int m) {
return ((x%m)+m)%m;
}
static int octaltodecimal(int deciNum)
{
int octalNum = 0, countval = 1;
int dNo = deciNum;
while (deciNum != 0) {
// decimals remainder is calculated
int remainder = deciNum % 8;
// storing the octalvalue
octalNum += remainder * countval;
// storing exponential value
countval = countval * 10;
deciNum /= 8;
}
return octalNum;
}
public static int floorSqrt(int x)
{
// Base Cases
if (x == 0 || x == 1)
return x;
// Do Binary Search for floor(sqrt(x))
int start = 1, end = x, ans=0;
while (start <= end)
{
int mid = (start + end) / 2;
// If x is a perfect square
if (mid*mid == x)
return mid;
// Since we need floor, we update answer when mid*mid is
// smaller than x, and move closer to sqrt(x)
if (mid*mid < x)
{
start = mid + 1;
ans = mid;
}
else // If mid*mid is greater than x
end = mid-1;
}
return ans;
}
static int div(int n,int b) {
int g=-1;
for(int i=2;i<=Math.sqrt(n);i++) {
if(n%i==0&&i!=b) {
g=i;
break;
}
}
return g;
}
public static long gcd(long a, long a2)
{
if (a == 0)
return a2;
return gcd(a2%a, a);
}
public static long lcm(long a, long b)
{
return (a*b)/gcd(a, b);
}
public static boolean isPrime(int n) {
if (n <= 1) {
return false;
}
if (n == 2) {
return true;
}
if (n % 2 == 0) {
return false;
}
for (int i = 3; i <= Math.sqrt(n) + 1; i = i + 2) {
if (n % i == 0) {
return false;
}
}
return true;
}
static class FastScanner {
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
StringTokenizer strToken = new StringTokenizer("");
String next() {
while (!strToken.hasMoreTokens()) {
try {
strToken = new StringTokenizer(input.readLine());
} catch (IOException e) {
e.printStackTrace();
}
} return strToken.nextToken();
}
String nextLine() {
String strrrrr = ""; try {
strrrrr = input.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return strrrrr;
}
int nextInt() {
return Integer.parseInt(next());
}
int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) a[i] = nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) a[i] = nextLong();
return a;
}
double nextDouble() {
return Double.parseDouble(next());
}
}
}
class DisjointSet {
private int[] parent;
private int[] rank;
public DisjointSet(int n) {
if (n < 0) throw new IllegalArgumentException();
parent = new int[n];
for(int i=0;i<n;i++) parent[i]=i;
rank = new int[n];
}
public int find(int x) {
if (parent[x] == x) return x;
return parent[x] = find(parent[x]); // Path compression by halving.
}
// Return false if x, y are connected.
public boolean union(int x, int y) {
int rootX = find(x);
int rootY = find(y);
if (rootX == rootY) return false;
// Make root of smaller rank point to root of larger rank.
if (rank[rootX] < rank[rootY]) parent[rootX] = rootY;
else if (rank[rootX] > rank[rootY]) parent[rootY] = rootX;
else {
parent[rootX] = rootY;
rank[rootY]++;
}
return true;
}
}
class Pair{
int value;
int distance;
Pair(int value,int distance){
this.value=value;
this.distance=distance;
}
}
class sortbydistance implements Comparator<Pair>{
@Override
public int compare(Pair o1, Pair o2) {
return o1.distance-o2.distance;
}
}
class Solution
{
static int[] dijkstra(int n, ArrayList<ArrayList<ArrayList<Integer>>> adj,ArrayList<Integer>ans)
{
PriorityQueue<Pair>set=new PriorityQueue<>(new sortbydistance());
int dis[]=new int[n];
Arrays.fill(dis,Integer.MAX_VALUE);
dis[0]=0;
while(set.isEmpty()==false) {
Pair p=set.poll();
int u=p.value;
for(int i=0;i<adj.get(u).size();i++) {
int v=adj.get(u).get(i).get(0);
int wt=adj.get(u).get(i).get(1);
if(dis[v]>dis[u]+wt) {
if(dis[v]!=Integer.MAX_VALUE) {
set.remove(new Pair(dis[v],v));
}
dis[v]=dis[u]+wt;
ans.add(v);
set.add(new Pair(v,dis[v]));
}
}
}
return dis;
}
}
/* Points-
*
* 1) when need max and min both use treeset if duplicates are there then use treeset<int []> and add a[1] as unique
* number but imp is to make new TreeSet<>((x,y)->(x[0]==y[0])?(x[1]-y[1]):(x[0]-y[0])) in this x[0]==y[0] cond.
* is important
*
* 2) n!/2 is not fac[n]/2 it is modmul(fac[n],inv[2],m)
*
* 3) if sorting can think of TreeSet also bro or want max and min use tree set as it has last and first function
*
* 4) if dp is not optimisable then think of changing the defination of dp states
*/
class pairs{
int a,b;
pairs(int a,int b){
this.a=a;this.b=b;
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 86aba5db2d089f0e535c6d15f6ab6f51 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.util.Scanner;
import java.util.*;
public class Main{
public static void main(String []args){
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-->0){
int n = scn.nextInt();
ArrayList<int[]> a = new ArrayList<>();
for(int i=0;i<n;i++){
a.add(new int[]{scn.nextInt(),i+1});
}
Collections.sort(a,(p1,p2)->p1[0]-p2[0]);
int c=0;
for(int i=0;i<n-1;i++){
for(int j=i+1;j<n&&(long)a.get(i)[0]*(long)a.get(j)[0]<=a.get(i)[1]+n;j++){
if(a.get(i)[0]*a.get(j)[0]==a.get(i)[1]+a.get(j)[1]) c++;
}
}
System.out.println(c);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 8f46a457ff7fd4996bdc5f0529b0400e | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Scanner;
import java.util.*;
public class Main{
// public static int floor(ArrayList<int[]> a,int i,int j,int m){
// while(i<=j){
// int mid = i+(j-i)/2;
// if(a.get(mid)[0]==m){
// return mid;
// }else if(a.get(mid)[0]<m){
// i=m+1;
// }else{
// j=m-1;
// }
// }
// return j;
// }
public static void main(String []args){
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-->0){
long n = scn.nextInt();
ArrayList<long[]> a = new ArrayList<>();
for(long i=1;i<=n;i++){
a.add(new long[]{scn.nextLong(),i});
}
Collections.sort(a,(p1,p2)->(int)(p1[0]-p2[0]));
int c=0;
for(int i=0;i<n-1;i++){
// int m = (2*n-1)/a.get(i)[0];
// int f = floor(a,i+1,n-1,m);
for(int j=i+1;j<n&&a.get(i)[0]*a.get(j)[0]<2*n;j++){
if(a.get(i)[0]*a.get(j)[0]==a.get(i)[1]+a.get(j)[1]) c++;
}
}
System.out.println(c);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | a2f0d5dc13cd569406991c3d41f93dfb | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Solution {
public static void main(String[] args) throws java.lang.Exception{
Scanner sc=new Scanner(System.in);
int T=sc.nextInt();
while(T-->0) {
int n=sc.nextInt();
int a[]=new int[n+1];
for(int i=1;i<=n;i++) {
a[i]=sc.nextInt();
}
long c=0;
for(int i=1;i<n;i++) {
for(int j=a[i]-i;j<=n;j+=a[i]) {
if(j>=1)
if(((long)a[i]*a[j]==i+j)&&(i<j))
c++;
}
}
System.out.println(c);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 23f85c95483655507226a19c7b917ebd | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.math.BigInteger;
import java.io.*;
public class Main implements Runnable {
public static void main(String[] args) {
new Thread(null, new Main(), "whatever", 1 << 26).start();
}
private FastScanner sc;
private PrintWriter pw;
public void run() {
try {
// pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
// sc = new FastScanner(new BufferedReader(new FileReader("input.txt")));
pw = new PrintWriter(System.out);
sc = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
} catch (Exception e) {
throw new RuntimeException();
}
int t = sc.nextInt();
// int t = 1;
while (t-- > 0) {
// sc.nextLine();
solve();
}
pw.close();
}
public long mod = 1_000_000_007;
public void solve() {
int n = sc.nextInt(), count = 0;
int[][] arr = new int[n][2];
for (int i = 0; i < n; i++) {
arr[i][1] = sc.nextInt();
arr[i][0] = i + 1;
}
Arrays.sort(arr, (o1, o2)->o1[1] - o2[1]);
long temp;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
temp = arr[i][1] * ((long)arr[j][1]);
if (temp >= 2 * n) break;
if (temp == arr[i][0] + arr[j][0]) {
count++;
}
}
}
pw.println(count);
}
class FastScanner {
private BufferedReader reader = null;
private StringTokenizer tokenizer = null;
public FastScanner(BufferedReader bf) {
reader = bf;
tokenizer = null;
}
public String next() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public String nextLine() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
return reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken("\n");
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public float nextFloat() {
return Float.parseFloat(next());
}
public int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String[] nextStringArray(int n) {
String[] a = new String[n];
for (int i = 0; i < n; i++) {
a[i] = next();
}
return a;
}
public long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = nextLong();
}
return a;
}
}
public long fastPow(long x, long y, long mod) {
if (y == 0) return 1;
if (y == 1) return x % mod;
long temp = fastPow(x, y / 2, mod);
long ans = (temp * temp) % mod;
return (y % 2 == 1) ? (ans * (x % mod)) % mod : ans;
}
// public long fastPow(long x, long y) {
// if (y == 0) return 1;
// if (y == 1) return x;
// long temp = fastPow(x, y / 2);
// long ans = (temp * temp);
// return (y % 2 == 1) ? (ans * x) : ans;
// }
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | d06eab4ed7a6d8c50c4aab3eddfc4c75 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.lang.*;
import java.math.BigInteger;
import java.io.*;
public class Main implements Runnable {
public static void main(String[] args) {
new Thread(null, new Main(), "whatever", 1 << 26).start();
}
private FastScanner sc;
private PrintWriter pw;
public void run() {
try {
// pw = new PrintWriter(new BufferedWriter(new FileWriter("output.txt")));
// sc = new FastScanner(new BufferedReader(new FileReader("input.txt")));
pw = new PrintWriter(System.out);
sc = new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
} catch (Exception e) {
throw new RuntimeException();
}
int t = sc.nextInt();
// int t = 1;
while (t-- > 0) {
// sc.nextLine();
solve();
}
pw.close();
}
public long mod = 1_000_000_007;
public void solve() {
int n = sc.nextInt(), count = 0;
int[] arr = new int[2 * n + 1];
Arrays.fill(arr, -1);
for (int i = 0; i < n; i++) {
arr[sc.nextInt()] = i + 1;
}
for (int i = 1; i <= 2 * n; i++) {
if (arr[i] == -1) continue;
for (int j = 1; j * i <= 2 * n; j++) {
if (i == j || arr[j] == -1) continue;
if (arr[i] + arr[j] == (long) i * j) {
++count;
}
}
}
count /= 2;
pw.println(count);
}
class FastScanner {
private BufferedReader reader = null;
private StringTokenizer tokenizer = null;
public FastScanner(BufferedReader bf) {
reader = bf;
tokenizer = null;
}
public String next() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public String nextLine() {
if (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
return reader.readLine();
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken("\n");
}
public long nextLong() {
return Long.parseLong(next());
}
public int nextInt() {
return Integer.parseInt(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
public float nextFloat() {
return Float.parseFloat(next());
}
public int[] nextIntArray(int n) {
int[] a = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String[] nextStringArray(int n) {
String[] a = new String[n];
for (int i = 0; i < n; i++) {
a[i] = next();
}
return a;
}
public long[] nextLongArray(int n) {
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = nextLong();
}
return a;
}
}
public long fastPow(long x, long y, long mod) {
if (y == 0) return 1;
if (y == 1) return x % mod;
long temp = fastPow(x, y / 2, mod);
long ans = (temp * temp) % mod;
return (y % 2 == 1) ? (ans * (x % mod)) % mod : ans;
}
// public long fastPow(long x, long y) {
// if (y == 0) return 1;
// if (y == 1) return x;
// long temp = fastPow(x, y / 2);
// long ans = (temp * temp);
// return (y % 2 == 1) ? (ans * x) : ans;
// }
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 752444fc804400ccc77b68633dd3458c | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class PlesentPairs {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int tt = sc.nextInt();
while(tt != 0){
int n = sc.nextInt();
ArrayList<Long> arr = new ArrayList<>();
for(int i = 0;i < n;i++){
arr.add(sc.nextLong());
}
int ans = 0;
for(int i = 0;i < arr.size();i++){
for(int j = (int) (arr.get(i)-i)-2;j < arr.size();j+=(arr.get(i))){
if(j > 0 ) {
if(i < j && arr.get(i) * arr.get(j) == (i + j + 2)) {
ans++;
}
}
}
}
System.out.println(ans);
tt--;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | a30f1c29da4faa69b46ed00ba78c7f42 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | /*import java.util.*;
public class PlesentPairs {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int tt = sc.nextInt();
while(tt != 0){
int n = sc.nextInt();
long arr[] = new long[n+5];
for(int i = 1;i <= n;i++){
arr[i] = sc.nextLong();
}
int ans = 0;
for(int i = 1;i <= n;i++){
for(int j = (int)(arr[i]-i);j <= n;j+=arr[i]){
if(j >= 0 ) {
if(i < j && arr[i] * arr[j] == i + j) {
ans++;
}
}
}
}
System.out.println(ans);
tt--;
}
}
}*/
import java.util.*;
public class PlesentPairs {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int tt = sc.nextInt();
while(tt != 0){
int n = sc.nextInt();
ArrayList<Long> arr = new ArrayList<>();
for(int i = 0;i < n;i++){
arr.add(sc.nextLong());
}
int ans = 0;
for(int i = 0;i < arr.size();i++){
for(int j = (int) (arr.get(i)-i)-2;j < arr.size();j+=(arr.get(i))){
if(j >= 0 && i < j && arr.get(i)*arr.get(j) == (i+j+2)){
ans++;
}
}
}
System.out.println(ans);
tt--;
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 20caeadfc6e00b379ced392c919643dc | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class PlesentPairs {
public static void main(String[] args){
Scanner sc = new Scanner(System.in);
int tt = sc.nextInt();
while(tt != 0){
int n = sc.nextInt();
long arr[] = new long[n+5];
for(int i = 1;i <= n;i++){
arr[i] = sc.nextLong();
}
int ans = 0;
for(int i = 1;i <= n;i++){
for(int j = (int)(arr[i]-i);j <= n;j+=arr[i]){
if(j >= 0 ) {
if(i < j && arr[i] * arr[j] == i + j) {
ans++;
}
}
}
}
System.out.println(ans);
tt--;
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 087fa15e74327c55da28e5f3f0f13f3a | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class pleasentpairs{
public static void main(String ...asda){
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int []a=new int[n+1];
for(int i=1;i<=n;i++)
a[i]=sc.nextInt();
int c=0;
for(int i=1;i<=n;i++){
int prod=a[i];
for(int j=prod;j<=2*n;j+=prod){
if(i<(j-i) && j-i<=n && j-i>0 && a[j-i]==j/a[i]){
c++;
//System.out.println(i+" "+j+" "+c);
}
if(j-i>n)
break;
}
//System.out.println("value "+i+" "+c);
}
System.out.println(c);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | fa503ddb250375a0edfca2119768c231 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
public class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main(String[] args) throws IOException
{
Reader s = new Reader();
int t = s.nextInt();
while(t-->0) {
int n = s.nextInt();
long[] a = new long[n+1];
for (int i = 1; i <= n; i++) {
a[i] = s.nextLong();
}
int ctr = 0;
for (int i = 1; i <= n; i++) {
for (int j = (int)a[i]-i; j <= n; j+=(int)a[i]) {
if(j>=0){
if(a[i]*a[j]==i+j && i<j)
ctr++;
}
}
}
System.out.println(ctr);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 6ecc312c62e3ab80b0ed9bbeab067658 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.DataInputStream;
import java.io.FileInputStream;
import java.io.IOException;
public class Main {
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main(String[] args) throws IOException
{
Reader s = new Reader();
int t = s.nextInt();
while(t-->0) {
int n = s.nextInt();
long[] a = new long[n];
for (int i = 0; i < n; i++) {
a[i] = s.nextLong();
}
int ctr = 0;
for (int i = 0; i < n; i++) {
int k = (int)(a[i]-2*(i+1)%a[i]);
for (int j = i+k; j < n; j+=(int)a[i]) {
if(a[i]*a[j]==i+j+2)
ctr++;
}
}
System.out.println(ctr);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 871b90940dacc77757fc7f8485900cdb | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class PleasantPairs {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(System.out);
static StringTokenizer st = new StringTokenizer("");
public static String next() {
try {
if (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
public static int ni() {
return Integer.parseInt(next());
}
public static void main(String[] args) throws IOException {
// try {
// br = new BufferedReader(new FileReader("C:/Users/gunjan.kumar01/Desktop/Java/codeforces/in.txt"));
// } catch (FileNotFoundException e) {
// e.printStackTrace();
// }
int t = ni();
while (t-- > 0) {
int n = ni();
// int[] arr = Arrays.stream(br.readLine().split("
// ")).mapToInt(Integer::parseInt).toArray();
int[] arr = new int[n + 1];
for (int i = 1; i < n + 1; i++) {
arr[i] = ni();
}
// solve(size, arr);
solve2(n, arr);
}
out.flush();
out.close();
}
public static void solve(int size, int[] arr) {
int less = 0;
// int more = 0;
int n = arr.length;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (i < j && (arr[i] * arr[j] == i + j + 2)) {
less++;
// out.println(" I < J : " + i + ", " + j + "\tPAIR : " + arr[i] + ", " +
// arr[j]);
} else {
// more++;
// out.println("NOT I < J : " + i + ", " + j + "\tPAIR : " + arr[i] + ", " +
// arr[j]);
}
}
}
// out.println(Arrays.toString(arr) + " | count total -" + (less + more) + "
// less : " + less + " more : " + more);
out.println(less);
}
public static void solve2(int n, int[] arr) {
int count = 0;
int len = arr.length;
int[] map = new int[2 * n + 2];
for (int x = 1; x < len; x++) {
map[arr[x]] = x;
}
for (int i = 1; i < len; i++) {
int x = arr[i]; // first number
int y = 1; // second number
int j = map[y]; // get index of second number
long prod = x * (long) y; // product
while (prod < 2 * n) {
if (j > i && prod == i + j)
count++;
// out.printf("First : %d Second : %d Index first : %d, Index second : %d ------
// Count %d \n", x, y,
// i + 1, j, count);
y++;
j = map[y];
prod = x * (long) y;
}
}
out.println(count);
}
public static void solve3(int n, int[] arr) {
int count = 0;
// int len = n+1;
for (int i = 1; i <= n; i++) {
int x = arr[i];
for (int j = x - i; j <= n; j += x) {
// int y = arr[j];
if (j > i && x * arr[j] == i + j) {
count++;
// out.printf("First : %d Second : %d Index first : %d, Index second : %d ------
// Count %d \n", x,
// arr[j], i , j, count);
}
}
}
out.println(count);
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | aa514a90d64eaac5cac3d54897b55c5d | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.StringTokenizer;
public class PleasantPairs {
static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
static PrintWriter out = new PrintWriter(System.out);
static StringTokenizer st = new StringTokenizer("");
public static String next() {
try {
if (st == null || !st.hasMoreTokens())
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
public static int ni() {
return Integer.parseInt(next());
}
public static void main(String[] args) throws IOException {
// try {
// br = new BufferedReader(new FileReader("C:/Users/gunjan.kumar01/Desktop/Java/codeforces/in.txt"));
// } catch (FileNotFoundException e) {
// e.printStackTrace();
// }
int t = ni();
while (t-- > 0) {
int n = ni();
// int[] arr = Arrays.stream(br.readLine().split("
// ")).mapToInt(Integer::parseInt).toArray();
int[] arr = new int[n + 1];
for (int i = 1; i < n + 1; i++) {
arr[i] = ni();
}
// solve(size, arr);
solve3(n, arr);
}
out.flush();
out.close();
}
public static void solve(int size, int[] arr) {
int less = 0;
// int more = 0;
int n = arr.length;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (i < j && (arr[i] * arr[j] == i + j + 2)) {
less++;
// out.println(" I < J : " + i + ", " + j + "\tPAIR : " + arr[i] + ", " +
// arr[j]);
} else {
// more++;
// out.println("NOT I < J : " + i + ", " + j + "\tPAIR : " + arr[i] + ", " +
// arr[j]);
}
}
}
// out.println(Arrays.toString(arr) + " | count total -" + (less + more) + "
// less : " + less + " more : " + more);
out.println(less);
}
public static void solve2(int n, int[] arr) {
int count = 0;
int len = arr.length;
int[] map = new int[2 * n + 1];
for (int x = 0; x < len; x++) {
map[arr[x]] = x + 1;
}
for (int i = 0; i < len; i++) {
int x = arr[i]; // first number
int y = 1; // second number
int j = map[y]; // get index of second number
long prod = x * y; // product
while (prod < 2 * n) {
if (j != 0 && prod == i + j + 1)
count++;
out.printf("First : %d Second : %d Index first : %d, Index second : %d ------ Count %d \n", x, y,
i + 1, j, count);
y++;
prod = x * y;
j = map[y];
}
}
out.println("Ans = " + count);
}
public static void solve3(int n, int[] arr) {
int count = 0;
// int len = n+1;
for (int i = 1; i <= n; i++) {
int x = arr[i];
for (int j = x - i; j <= n; j += x) {
// int y = arr[j];
if (j > i && x * (long)arr[j] == i + j) {
count++;
// out.printf("First : %d Second : %d Index first : %d, Index second : %d ------ Count %d \n", x,
// arr[j], i , j, count);
}
}
}
out.println(count);
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | a572ef33e63c7b45cf98f89ec36901fd | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class Extras {
static class pair {
long first;
long second;
pair(long f, long s) {
first = f;
second = s;
}
// @Override
// public String toString() {
// return "pair{" +
// "first=" + first +
// ", second=" + second +
// '}';
// }
}
public static void main(final String[] args) throws IOException {
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
while (t-- > 0) {
int n = Integer.parseInt(br.readLine());
String[] inp = br.readLine().split(" ");
pair[] a = new pair[n];
for(int i = 0; i < n; i++) {
long input = Long.parseLong(inp[i]);
a[i] = new pair(input, i+1);
}
Arrays.sort(a, Comparator.comparing((pair x) -> x.first));
// System.out.println(Arrays.toString(a));
int count = 0;
for(int i = 0; i < n - 1; i++) {
for(int j = i + 1; j < n; j++) {
// System.out.println(i + ", " + j);
if(a[i].first * a[j].first >= 2L * n) break;
if(a[i].first * a[j].first == a[i].second + a[j].second) {
count += 1;
}
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 441412cade27a80dead93256b1267b0c | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual solution is at the top
*/
public class Main {
public static void main(String[] args) {
InputStream inputStream = System.in;
OutputStream outputStream = System.out;
InputReader in = new InputReader(inputStream);
PrintWriter out = new PrintWriter(outputStream);
TaskB solver = new TaskB();
int testCount = Integer.parseInt(in.next());
for (int i = 1; i <= testCount; i++)
solver.solve(i, in, out);
out.close();
}
static class TaskB {
public void solve(int testNumber, InputReader in, PrintWriter out) {
long n = in.nextInt();
long[] arr = new long[(int) (n + 1)]; //i 가 1부터 시작해야 편하므로,
long count = 0;
for (long i = 1; i <= n; i++) {
arr[(int) i] = in.nextInt();
}
for (long i = 1; i <= n - 1; i++) {
for (long j = arr[(int) i] - i; j <= n; j += arr[(int) i]) {
if (j > 1) {
//ai - i 가 음수이거나 0이 나올 수 있다
if (arr[(int) i] * arr[(int) j] == i + j && i < j) {
//i + j = ai * aj && j가 i보다 커야한다
count++;
}
}
}
}
out.println(count);
}
}
static class InputReader {
public BufferedReader reader;
public StringTokenizer tokenizer;
public InputReader(InputStream stream) {
reader = new BufferedReader(new InputStreamReader(stream), 32768);
tokenizer = null;
}
public String next() {
while (tokenizer == null || !tokenizer.hasMoreTokens()) {
try {
tokenizer = new StringTokenizer(reader.readLine());
} catch (IOException e) {
throw new RuntimeException(e);
}
}
return tokenizer.nextToken();
}
public int nextInt() {
return Integer.parseInt(next());
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 960c5fd18d738aed0851ef64003827cb | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.math.*;
import java.util.*;
/*
author : Multi-Thread
*/
public class B {
// public class Main {
// static int INF = 998244353;
static int INF = (int) 1e9 + 7;
static int MAX = Integer.MAX_VALUE;
static int MIN = Integer.MIN_VALUE;
public static void main(String[] args) {
int test = fs.nextInt();
// int test = 1;
for (int cases = 0; cases < test; cases++) {
// solve();
System.out.print(solve() + "\n");
}
}
static String solve() {
int n = fs.nextInt();
long[] hm = new long[2 * n + 1];
long[] rev = new long[n + 1];
ArrayList<Long> al = new ArrayList<Long>();
for (int i = 0; i < n; i++) {
long x = fs.nextLong();
hm[(int) x] = (long) i + 1;
rev[i + 1] = (long) x;
al.add(x);
}
Collections.sort(al);
long ans = 0;
for (int i = 0; i < n; i++) {
long index = hm[al.get(i).intValue()];
long j = al.get(i) + 1;
for (; j * al.get(i) <= n + n - 1; j++) {
long need = (long) al.get(i) * j;
long index2 = need - index;
if (index2 >= 1 && index2 <= n) {
long next = rev[(int) index2];
if (next == j) {
++ans;
}
}
}
}
return ans + "";
}
static void intSort(int[] a, boolean reverse) {
ArrayList<Integer> al = new ArrayList<Integer>();
for (int i : a)
al.add(i);
Collections.sort(al);
if (reverse) {
for (int i = 0; i < a.length; i++)
a[i] = al.get(a.length - i - 1);
} else {
for (int i = 0; i < a.length; i++)
a[i] = al.get(i);
}
}
static void longSort(long[] a, boolean reverse) {
ArrayList<Long> al = new ArrayList<>();
for (long i : a)
al.add(i);
Collections.sort(al);
if (reverse) {
for (int i = 0; i < a.length; i++)
a[i] = al.get(a.length - i - 1);
} else {
for (int i = 0; i < a.length; i++)
a[i] = al.get(i);
}
}
public static int[] radixSort(int[] f) {
return radixSort(f, f.length);
}
public static int[] radixSort(int[] f, int n) {
int[] to = new int[n];
{
int[] b = new int[65537];
for (int i = 0; i < n; i++)
b[1 + (f[i] & 0xffff)]++;
for (int i = 1; i <= 65536; i++)
b[i] += b[i - 1];
for (int i = 0; i < n; i++)
to[b[f[i] & 0xffff]++] = f[i];
int[] d = f;
f = to;
to = d;
}
{
int[] b = new int[65537];
for (int i = 0; i < n; i++)
b[1 + (f[i] >>> 16)]++;
for (int i = 1; i <= 65536; i++)
b[i] += b[i - 1];
for (int i = 0; i < n; i++)
to[b[f[i] >>> 16]++] = f[i];
int[] d = f;
f = to;
to = d;
}
return f;
}
static class Pair {
int first, second;
public Pair(int first, int second) {
this.first = first;
this.second = second;
}
public String toString() {
return "[" + first + "," + second + "]";
}
}
static class LongPair {
long first;
long second;
LongPair(long a, long b) {
this.first = a;
this.second = b;
}
public String toString() {
return "[" + first + "," + second + "]";
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static class OutputWriter {
private final PrintWriter writer;
public OutputWriter(OutputStream outputStream) {
writer = new PrintWriter(new BufferedWriter(new OutputStreamWriter(outputStream)));
}
public OutputWriter(Writer writer) {
this.writer = new PrintWriter(writer);
}
public void print(Object... objects) {
for (int i = 0; i < objects.length; i++) {
if (i != 0)
writer.print(' ');
writer.print(objects[i]);
}
}
public void println() {
writer.print("\n");
}
public void printLine(Object... objects) {
print(objects);
writer.println();
}
public void close() {
writer.close();
}
public void flush() {
writer.flush();
}
}
private static final FastReader fs = new FastReader();
private static final OutputWriter out = new OutputWriter(System.out);
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 7c528ba26b4794e72ff3014787b77a24 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Scanner;
import java.io.OutputStreamWriter;
import java.io.BufferedWriter;
import java.io.IOException;
public class pleasantpairs {
public static void main(String[] args) throws IOException {
Scanner in = new Scanner(System.in);
int tt = in.nextInt();
BufferedWriter log = new BufferedWriter(new OutputStreamWriter(System.out));
for(int w = 0; w<tt; w++) {
int n = in.nextInt();
int[] a = new int[n+1];
for(int i = 1; i<=n; i++) {
a[i] = in.nextInt();
}
int broj = 0;
for(int i = 1; i<=n; i++) {
for(int j = a[i]-i; j<=n; j+=a[i]) {
if(j<=i) {
continue;
}
if((long) a[i]*a[j]==i+j) {
broj++;
}
}
}
log.write(broj+"\n");
log.flush();
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | c2dd14ca3b2abd57f2000cbf00c6e988 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.io.*;
import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.*;
import static java.lang.System.out ;
public class CP0065 {
public static void main(String args[])throws Exception{
PrintWriter pw = new PrintWriter(out);
FastReader sc = new FastReader();
Reader s = new Reader();
int test;
//code yaha se hai
try{
test = s.nextInt();
}
catch(Exception E){
return;
}
while (test-- > 0){
int n = s.nextInt();
HashMap<Integer , Integer> mp = new HashMap<>();
ArrayList<Integer> ar = new ArrayList<>();
for (int i = 0; i < n; i++) {
ar.add(s.nextInt());
mp.put(ar.get(i) , i+1);
}
Collections.sort(ar);
long val ;
int count = 0 ;
for (int i = 0; i < n; i++) {
if(ar.get(i) >= 2 * n){
break;
}
for (int j = i+1; j < n; j++) {
val = 1l * ar.get(i)*ar.get(j) ;
if(val >= 2 * n ){
break;
}
if(val == (mp.get(ar.get(i)) + mp.get(ar.get(j)))){
count++;
}
}
}
pw.println(count);
}
pw.close();
}
/*
Note :
Try using sorting , when approach is of O(N^2)
As sorting takes : O( N Log N )
*/
/*
len : Length of Array / inputs
num : key to map
//Putting values in map <Integer , ArrayList>
for (int i = 0; i < len; i++) {
num =sc.nextInt();
if(!mp.containsKey(num)){
mp.put(num ,new ArrayList<>());
}
}
//getting size of ArrayList
for(Integer key : mp.keySet()){
maxValue = max(maxValue , mp.get(key).size());
}
*/
/*
Sieve
boolean[] bool = new boolean[num];
for (int i = 0; i< bool.length; i++) {
bool[i] = true;
}
for (int i = 2; i< Math.sqrt(num); i++) {
if(bool[i] == true) {
for(int j = (i*i); j<num; j = j+i) {
bool[j] = false;
}
}
}
*/
static long exponentiation(long base, long exp)
{
long N = 1000000007 ;
if (exp == 0)
return 1;
if (exp == 1)
return base % N;
long t = exponentiation(base, exp / 2);
t = (t * t) % N;
// if exponent is even value
if (exp % 2 == 0)
return t;
// if exponent is odd value
else
return ((base % N) * t) % N;
}
public static int BinarySearch (long ar[] , int start , int end , long key , boolean lower){
int mid ;
boolean found = false;
int ans = 0;
if(lower){
while(start <= end){
mid = start + (end-start) / 2;
if(ar[mid] < key ){
if(!found){
ans = mid;
}
start = mid+1;
}
else{
if(ar[mid] == key){
found = true;
ans = mid ;
}
end = mid-1;
}
}
}
else{
while(start <= end){
mid = start + (end-start) / 2;
if(ar[mid] <= key ){
if(ar[mid] == key){
found = true;
ans = mid;
}
start = mid+1;
}
else{
if(!found){
ans = mid;
}
end = mid-1;
}
}
}
return ans;
}
public static int BinarySearchArrayList (ArrayList<Long> ar , int start , int end , long key , boolean lower){
int mid ;
boolean found = false;
int ans = -1;
if(lower){
while(start <= end){
mid = start + (end-start) / 2;
if(ar.get(mid) < key ){
if(!found){
ans = mid;
}
start = mid+1;
}
else{
if(ar.get(mid) == key){
found = true;
ans = mid ;
}
end = mid-1;
}
}
}
else{
while(start <= end){
mid = start + (end-start) / 2;
if(ar.get(mid) <= key ){
if(ar.get(mid) == key){
found = true;
ans = mid;
}
start = mid+1;
}
else{
if(!found){
ans = mid;
}
end = mid-1;
}
}
}
return ans;
}
public static String reverseString(String str){
StringBuilder input1 = new StringBuilder();
// append a string into StringBuilder input1
input1.append(str);
// reverse StringBuilder input1
input1.reverse();
return input1+"";
}
public static void sort(int[] arr){
//because Arrays.sort() uses quick sort , Worst Complexity : o(N^2)
ArrayList <Integer> ls = new ArrayList<>();
for(int x:arr){
ls.add(x);
}
Collections.sort(ls);
//Collections.sort(arrayList) : Uses Merge Sort : Complexity O(NlogN)
for(int i = 0 ; i < arr.length ; i++){
arr[i] = ls.get(i);
}
}
public static long gcd(long a , long b){
if(a>b){
a = (a+b) - (a = b);
}
if(a == 0L){
return b;
}
return gcd(b%a ,a);
}
public static boolean isPrime(long n){
if(n < 2){
return false;
}
if(n == 2 || n == 3){
return true;
}
if(n % 2 == 0 || n % 3 == 0){
return false;
}
long sq = (long)Math.sqrt(n) + 1;
for(long i = 5L; i <=sq ; i=i+6){
if(n % i == 0 || n % (i+2) == 0){
return false;
}
}
return true;
}
public static class Pair implements Comparable<Pair> {
long first ;
long second;
Pair(long fst , long scnd){
first = fst;
second = scnd;
}
public int compareTo(Pair obj){
if(this.first > obj.first){
return 1;
}
else if(this.first < obj.first){
return -1;
}
return 0;
}
}
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while(st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch(IOException e){
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 167c29856c1386f9677fc66215d22084 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | // import static java.lang.System.out;
import static java.lang.Math.*;
import java.awt.desktop.AboutHandler;
import java.lang.reflect.Array;
import java.util.*;
import java.io.*;
import java.math.*;
import java.util.concurrent.Phaser;
public class Main {
static FastReader sc;
static long mod = ((long) 1e9) + 7;
// static FastWriter out;
static Tree tree;
static PrintWriter out, err;
static int imax = Integer.MAX_VALUE;
static int imin = Integer.MIN_VALUE;
static long lmax = Long.MAX_VALUE;
static long lmin = Long.MIN_VALUE;
static double dmax = Double.MAX_VALUE;
static double dmin = Double.MIN_VALUE;
public static void main(String hi[]) throws IOException {
// initializeIO();
out = new PrintWriter(System.out);
err = new PrintWriter(System.err);
sc = new FastReader();
// sc=new Scanner(System.in);
long startTimeProg = System.currentTimeMillis();
long endTimeProg;
int t = 1;
t = sc.nextInt();
// boolean[] seave=sieveOfEratosthenes((int)(1e5));
// int[] seave2=sieveOfEratosthenesInt((long)sqrt(1e9));
// debug(seave2);
while (t-- != 0) {
int n = sc.nextInt();
long[][] arr = new long[n][2];
for (int i = 0; i < n; i++) {
arr[i][0] = sc.nextInt();
arr[i][1] = i + 1;
}
Arrays.sort(arr,(a,b)->{
return (int)(a[0]-b[0]);
});
debug(arr);
long size=(2*n)-1;
long c = 0;
for (int i = 0; i < n; i++) {
for (int j = i + 1; j < n; j++) {
if (arr[i][0] * arr[j][0] > size) break;
if (arr[i][0] * arr[j][0] == arr[i][1] + arr[j][1]) c++;
}
}
print(c);
}
endTimeProg = System.currentTimeMillis();
debug("[finished : " + (endTimeProg - startTimeProg) + ".ms ]");
out.flush();
err.flush();
// System.out.println(String.format("%.9f", max));
}
private static long calculateSum(char[] s) throws IOException {
int n=s.length;
long ans=0;
for(int i =0;i<n-1;i++){
String st=((s[i]-'0')+""+(s[i+1]-'0'));
// debug(st);
ans+=Integer.parseInt(st);
}
return ans;
}
private static boolean chk(long[] res) {
int n=res.length;
for(int i=0;i<n;i++){
if((res[(i+1)%n]<res[i]&&res[((i-1)<0)?n-1:i-1]<res[i])||
(res[(i+1)%n]>res[i]&&res[((i-1)<0)?n-1:i-1]>res[i]))continue;
else return false;
}
return true;
}
static String game(int[] arr1,int[] arr2,boolean see){
int n=arr1.length;
int m=arr2.length;
int i=n-1,j=m-1;
int card=0;
while (i>=0&&j>=0){
if(see){
while (i>=0&&arr1[i]<=card)i--;
if(i<0){
return "Bob";
}
card=arr1[i];
}else{
while (j>=0&&arr2[j]<=card)j--;
if(j<0){
return "Alice";
}
card=arr2[j];
}
see=!see;
}
return "alice";
}
static int CeilIndex(int A[], int l, int r, int key)
{
while (r - l > 1) {
int m = l + (r - l) / 2;
if (A[m] > key)
r = m;
else
l = m;
}
return r;
}
static int LongestIncreasingSubsequenceLength(int A[], int size)
{
// Add boundary case, when array size is one
int[] tailTable = new int[size];
int len; // always points empty slot
tailTable[0] = A[0];
len = 1;
for (int i = 1; i < size; i++) {
if (A[i] < tailTable[0])
// new smallest value
tailTable[0] = A[i];
else if (A[i] > tailTable[len - 1])
// A[i] wants to extend largest subsequence
tailTable[len++] = A[i];
else
// A[i] wants to be current end candidate of an existing
// subsequence. It will replace ceil value in tailTable
tailTable[CeilIndex(tailTable, -1, len - 1, A[i])] = A[i];
}
return len;
}
public static int log2(int N)
{
// calculate log2 N indirectly
// using log() method
int result = (int)(Math.log(N) / Math.log(2));
return result;
}
private static String toString(int[] arr){
StringBuilder sb=new StringBuilder();
for(int i=0;i<arr.length;i++)sb.append(arr[i]+" ");
return sb.toString();
}
private static String toString(long[] arr){
StringBuilder sb=new StringBuilder();
for(int i=0;i<arr.length;i++)sb.append(arr[i]+" ");
return sb.toString();
}
private static String toString(List<Integer> arr){
StringBuilder sb=new StringBuilder();
for(int i=0;i<arr.size();i++)sb.append(arr.get(i)+" ");
return sb.toString();
}
private static String toStringLong(List<Long> arr){
StringBuilder sb=new StringBuilder();
for(int i=0;i<arr.size();i++)sb.append(arr.get(i)+" ");
return sb.toString();
}
private static void print(String s) throws IOException {
out.println(s);
}
static boolean isSubsetSum(int set[], int n, int sum) {
// The value of subset[i][j] will be
// true if there is a subset of
// set[0..j-1] with sum equal to i
boolean subset[][] = new boolean[sum + 1][n + 1];
// If sum is 0, then answer is true
for (int i = 0; i <= n; i++)
subset[0][i] = true;
// If sum is not 0 and set is empty,
// then answer is false
for (int i = 1; i <= sum; i++)
subset[i][0] = false;
// Fill the subset table in bottom
// up manner
for (int i = 1; i <= sum; i++) {
for (int j = 1; j <= n; j++) {
subset[i][j] = subset[i][j - 1];
if (i >= set[j - 1])
subset[i][j] = subset[i][j]
|| subset[i - set[j - 1]][j - 1];
}
}
return subset[sum][n];
}
private static String chk(long[] arr, int n) throws IOException {
boolean see=true;
int dec=0,eq=0;
int s=0;
for(int i=1;i<n;i++){
if(arr[i]<=arr[i-1]) {
if(!see&&arr[i]<arr[i-1])return "no";
if(arr[i]<arr[i-1]&&s<=0)return "no";
s--;
}else{
if(s>1)return "no";
see=false;
}
}
return "yes";
}
private static boolean isPeak(int[] arr,int i,int l,int r){
if(i==l||i==r)return false;
return (arr[i+1]<arr[i]&&arr[i]>arr[i-1]);
}
private static long kadens(List<Long> li, int l, int r) {
long max = Long.MIN_VALUE;
long ans = 0;
for (int i = l; i <= r; i++) {
ans += li.get(i);
max = max(ans, max);
if (ans < 0) ans = 0;
}
return max;
}
public static boolean isNumeric(String strNum) {
if (strNum == null) {
return false;
}
try {
double d = Double.parseDouble(strNum);
} catch (NumberFormatException nfe) {
return false;
}
return true;
}
private static boolean isSorted(List<Integer> li) {
int n = li.size();
if (n <= 1) return true;
for (int i = 0; i < n - 1; i++) {
if (li.get(i) > li.get(i + 1)) return false;
}
return true;
}
private static boolean isSorted(int[] arr) {
int n = arr.length;
if (n <= 1) return true;
for (int i = 0; i < n - 1; i++) {
if (arr[i] >arr[i+1]) return false;
}
return true;
}
static boolean isPowerOfTwo(long x) {
return x != 0 && ((x & (x - 1)) == 0);
}
private static boolean ispallindromeList(List<Integer> res) {
int l = 0, r = res.size() - 1;
while (l < r) {
if (res.get(l) != res.get(r)) return false;
l++;
r--;
}
return true;
}
private static class Pair {
int first = 0;
int sec = 0;
int[] arr;
char ch;
String s;
Map<Integer, Integer> map;
Pair(int first, int sec) {
this.first = first;
this.sec = sec;
}
Pair(int[] arr) {
this.map = new HashMap<>();
for (int x : arr) this.map.put(x, map.getOrDefault(x, 0) + 1);
this.arr = arr;
}
Pair(char ch, int first) {
this.ch = ch;
this.first = first;
}
Pair(String s, int first) {
this.s = s;
this.first = first;
}
}
private static int sizeOfSubstring(int st, int e) {
int s = e - st + 1;
return (s * (s + 1)) / 2;
}
private static Set<Long> factors(long n) {
Set<Long> res = new HashSet<>();
// res.add(n);
for (long i = 1; i * i <= (n); i++) {
if (n % i == 0) {
res.add(i);
if (n / i != i) {
res.add(n / i);
}
}
}
return res;
}
private static long fact(long n) {
if (n <= 2) return n;
return n * fact(n - 1);
}
private static long ncr(long n, long r) {
return fact(n) / (fact(r) * fact(n - r));
}
private static int lessThen(long[] nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums[mid] <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int lessThen(List<Long> nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int lessThen(List<Integer> nums, int val, boolean work, int l, int r) {
int i = -1;
if (work) i = 0;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) <= val) {
i = mid;
l = mid + 1;
} else {
r = mid - 1;
}
}
return i;
}
private static int greaterThen(List<Long> nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static int greaterThen(List<Integer> nums, int val, boolean work) {
int i = -1, l = 0, r = nums.size() - 1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums.get(mid) >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static int greaterThen(long[] nums, long val, boolean work, int l, int r) {
int i = -1;
if (work) i = r;
while (l <= r) {
int mid = l + ((r - l) / 2);
if (nums[mid] >= val) {
i = mid;
r = mid - 1;
} else {
l = mid + 1;
}
}
return i;
}
private static void swap(char[] s,int i,int j){
char t=s[i];
s[i]=s[j];
s[j]=t;
}
private static long gcd(long[] arr) {
long ans = 0;
for (long x : arr) {
ans = gcd(x, ans);
}
return ans;
}
private static int gcd(int[] arr) {
int ans = 0;
for (int x : arr) {
ans = gcd(x, ans);
}
return ans;
}
private static long sumOfAp(long a, long n, long d) {
long val = (n * (2 * a + ((n - 1) * d)));
return val / 2;
}
//geometrics
private static double areaOftriangle(double x1, double y1, double x2, double y2, double x3, double y3) {
double[] mid_point = midOfaLine(x1, y1, x2, y2);
// debug(Arrays.toString(mid_point)+" "+x1+" "+y1+" "+x2+" "+y2+" "+x3+" "+" "+y3);
double height = distanceBetweenPoints(mid_point[0], mid_point[1], x3, y3);
double wight = distanceBetweenPoints(x1, y1, x2, y2);
// debug(height+" "+wight);
return (height * wight) / 2;
}
private static double distanceBetweenPoints(double x1, double y1, double x2, double y2) {
double x = x2 - x1;
double y = y2 - y1;
return (Math.pow(x, 2) + Math.pow(y, 2));
}
public static boolean isPerfectSquareByUsingSqrt(long n) {
if (n <= 0) {
return false;
}
double squareRoot = Math.sqrt(n);
long tst = (long) (squareRoot + 0.5);
return tst * tst == n;
}
private static double[] midOfaLine(double x1, double y1, double x2, double y2) {
double[] mid = new double[2];
mid[0] = (x1 + x2) / 2;
mid[1] = (y1 + y2) / 2;
return mid;
}
private static long sumOfN(long n) {
return (n * (n + 1)) / 2;
}
private static long power(long x, long y, long p) {
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0) {
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
/* Function to calculate x raised to the power y in O(logn)*/
static long power(long x, long y) {
long temp;
if (y == 0)
return 1l;
temp = power(x, y / 2);
if (y % 2 == 0)
return (temp * temp);
else
return (x * temp * temp);
}
private static StringBuilder reverseString(String s) {
StringBuilder sb = new StringBuilder(s);
int l = 0, r = sb.length() - 1;
while (l <= r) {
char ch = sb.charAt(l);
sb.setCharAt(l, sb.charAt(r));
sb.setCharAt(r, ch);
l++;
r--;
}
return sb;
}
private static void swap(List<Integer> li, int i, int j) {
int t = li.get(i);
li.set(i, li.get(j));
li.set(j, t);
}
private static void swap(int[] arr, int i, int j) {
int t = arr[i];
arr[i]=arr[j];
arr[j]=t;
}
static int lcm(int a, int b) {
return (a / gcd(a, b)) * b;
}
static long lcm(long a, long b) {
return (a / gcd(a, b)) * b;
}
private static String decimalToString(int x) {
return Integer.toBinaryString(x);
}
private static String decimalToString(long x) {
return Long.toBinaryString(x);
}
private static boolean isPallindrome(String s, int l, int r) {
while (l < r) {
if (s.charAt(l) != s.charAt(r)) return false;
l++;
r--;
}
return true;
}
private static StringBuilder removeLeadingZero(StringBuilder sb) {
int i = 0;
while (i < sb.length() && sb.charAt(i) == '0') i++;
// debug("remove "+i);
if (i == sb.length()) return new StringBuilder();
return new StringBuilder(sb.substring(i, sb.length()));
}
private static StringBuilder removeEndingZero(StringBuilder sb) {
int i = sb.length() - 1;
while (i >= 0 && sb.charAt(i) == '0') i--;
if (i < 0) return new StringBuilder();
return new StringBuilder(sb.substring(0, i + 1));
}
private static int stringToDecimal(String binaryString) {
return Integer.parseInt(binaryString, 2);
}
private static void debug(int[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(long[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(List<int[]> arr) {
for (int[] a : arr) {
err.println(Arrays.toString(a));
}
}
private static void debug(float[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(double[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(boolean[][] arr) {
for (int i = 0; i < arr.length; i++) {
err.println(Arrays.toString(arr[i]));
}
}
// private static void print() throws IOException {
// out.println(s);
// }
private static void debug(String s) throws IOException {
err.println(s);
}
private static int charToIntS(char c) {
return ((((int) (c - '0')) % 48));
}
private static void print(double s) throws IOException {
out.println(s);
}
private static void print(float s) throws IOException {
out.println(s);
}
private static void print(long s) throws IOException {
out.println(s);
}
private static void print(int s) throws IOException {
out.println(s);
}
private static void debug(double s) throws IOException {
err.println(s);
}
private static void debug(float s) throws IOException {
err.println(s);
}
private static void debug(long s) {
err.println(s);
}
private static void debug(int s) {
err.println(s);
}
private static boolean isPrime(long n) {
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i * i <= n; i += 2) {
if (n % i == 0)
return false;
}
return true;
}
private static List<List<Integer>> readUndirectedGraph(int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < n; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readUndirectedGraph(int[][] intervals, int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < intervals.length; i++) {
int x = intervals[i][0];
int y = intervals[i][1];
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int[][] intervals, int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < intervals.length; i++) {
int x = intervals[i][0];
int y = intervals[i][1];
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int n) {
List<List<Integer>> graph = new ArrayList<>();
for (int i = 0; i <= n; i++) {
graph.add(new ArrayList<>());
}
for (int i = 0; i < n; i++) {
int x = sc.nextInt();
int y = sc.nextInt();
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
static String[] readStringArray(int n) {
String[] arr = new String[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.next();
}
return arr;
}
private static Map<Character, Integer> freq(String s) {
Map<Character, Integer> map = new HashMap<>();
for (char c : s.toCharArray()) {
map.put(c, map.getOrDefault(c, 0) + 1);
}
return map;
}
private static Map<Long, Integer> freq(long[] arr) {
Map<Long, Integer> map = new HashMap<>();
for (long x : arr) {
map.put(x, map.getOrDefault(x, 0) + 1);
}
return map;
}
private static Map<Integer, Integer> freq(int[] arr) {
Map<Integer, Integer> map = new HashMap<>();
for (int x : arr) {
map.put(x, map.getOrDefault(x, 0) + 1);
}
return map;
}
static boolean[] sieveOfEratosthenes(long n) {
boolean prime[] = new boolean[(int) n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++) {
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true) {
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static int[] sieveOfEratosthenesInt(long n) {
boolean prime[] = new boolean[(int) n + 1];
Set<Integer> li = new HashSet<>();
for (int i = 2; i <= n; i++) {
prime[i] = true;
}
for (int p = 2; p * p <= n; p++) {
if (prime[p] == true) {
for (int i = p * p; i <= n; i += p) {
prime[i] = false;
}
}
}
for(int i=0;i<=n;i++){
if(prime[i])li.add(i);
}
int[] arr = new int[li.size()];
int i = 0;
for (int x : li) {
arr[i++] = x;
}
return arr;
}
public static long Kadens(List<Long> prices) {
long sofar = 0;
long max_v = 0;
for (int i = 0; i < prices.size(); i++) {
sofar += prices.get(i);
if (sofar < 0) {
sofar = 0;
}
max_v = Math.max(max_v, sofar);
}
return max_v;
}
public static int Kadens(int[] prices) {
int sofar = 0;
int max_v = 0;
for (int i = 0; i < prices.length; i++) {
sofar += prices[i];
if (sofar < 0) {
sofar = 0;
}
max_v = Math.max(max_v, sofar);
}
return max_v;
}
static boolean isMemberAC(int a, int d, int x) {
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
static boolean isMemberAC(long a, long d, long x) {
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
private static void sort(int[] arr) {
int n = arr.length;
List<Integer> li = new ArrayList<>();
for (int x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(int[] arr) {
int n = arr.length;
List<Integer> li = new ArrayList<>();
for (int x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sort(double[] arr) {
int n = arr.length;
List<Double> li = new ArrayList<>();
for (double x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(double[] arr) {
int n = arr.length;
List<Double> li = new ArrayList<>();
for (double x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sortReverse(long[] arr) {
int n = arr.length;
List<Long> li = new ArrayList<>();
for (long x : arr) {
li.add(x);
}
Collections.sort(li, Collections.reverseOrder());
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static void sort(long[] arr) {
int n = arr.length;
List<Long> li = new ArrayList<>();
for (long x : arr) {
li.add(x);
}
Collections.sort(li);
for (int i = 0; i < n; i++) {
arr[i] = li.get(i);
}
}
private static long sum(int[] arr) {
long sum = 0;
for (int x : arr) {
sum += x;
}
return sum;
}
private static long sum(long[] arr) {
long sum = 0;
for (long x : arr) {
sum += x;
}
return sum;
}
private static long evenSumFibo(long n) {
long l1 = 0, l2 = 2;
long sum = 0;
while (l2 < n) {
long l3 = (4 * l2) + l1;
sum += l2;
if (l3 > n) break;
l1 = l2;
l2 = l3;
}
return sum;
}
private static void initializeIO() {
try {
System.setIn(new FileInputStream("input"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
System.setErr(new PrintStream(new FileOutputStream("error.txt")));
} catch (Exception e) {
System.err.println(e.getMessage());
}
}
private static int maxOfArray(int[] arr) {
int max = Integer.MIN_VALUE;
for (int x : arr) {
max = Math.max(max, x);
}
return max;
}
private static long maxOfArray(long[] arr) {
long max = Long.MIN_VALUE;
for (long x : arr) {
max = Math.max(max, x);
}
return max;
}
private static int[][] readIntIntervals(int n, int m) {
int[][] arr = new int[n][m];
for (int j = 0; j < n; j++) {
for (int i = 0; i < m; i++) {
arr[j][i] = sc.nextInt();
}
}
return arr;
}
private static long gcd(long a, long b) {
if (b == 0) return a;
return gcd(b, a % b);
}
private static int gcd(int a, int b) {
if (b == 0) return a;
return gcd(b, a % b);
}
private static int[] readIntArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextInt();
}
return arr;
}
private static double[] readDoubleArray(int n) {
double[] arr = new double[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextDouble();
}
return arr;
}
private static long[] readLongArray(int n) {
long[] arr = new long[n];
for (int i = 0; i < n; i++) {
arr[i] = sc.nextLong();
}
return arr;
}
private static void print(int[] arr) throws IOException {
out.println(Arrays.toString(arr));
}
private static void print(long[] arr) throws IOException {
out.println(Arrays.toString(arr));
}
private static void print(String[] arr) throws IOException {
out.println(Arrays.toString(arr));
}
private static void print(double[] arr) throws IOException {
out.println(Arrays.toString(arr));
}
private static void debug(String[] arr) {
err.println(Arrays.toString(arr));
}
private static void debug(Boolean[][] arr) {
for (int i = 0; i < arr.length; i++) err.println(Arrays.toString(arr[i]));
}
private static void debug(int[] arr) {
err.println(Arrays.toString(arr));
}
private static void debug(long[] arr) {
err.println(Arrays.toString(arr));
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class Tree{
List<List<Integer>> adj=new ArrayList<>();
int parent=1;
int[] arr;
int n;
Map<Integer,Integer> node_parent;
List<Integer> leaf;
public Tree(int n) {
this.n=n;
leaf=new ArrayList<>();
node_parent=new HashMap<>();
arr=new int[n+2];
for(int i=0;i<=n+1;i++){
adj.add(new ArrayList<>());
}
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
node_parent.put(i+1,arr[i]);
if((i+1)==arr[i])parent=arr[i];
else adj.get(arr[i]).add(i+1);
}
}
private List<List<Integer>> getTree(){
return adj;
}
private void formLeaf(int v){
boolean leaf=true;
for(int x:adj.get(v)){
formLeaf(x);
leaf=false;
}
if(leaf)this.leaf.add(v);
}
private List<Integer> getLeaf(){
return leaf;
}
private Map<Integer,Integer> getParents(){
return node_parent;
}
int[] getArr(){
return arr;
}
}
static class Dsu {
int[] parent, size;
Dsu(int n) {
parent = new int[n + 1];
size = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
size[i] = 1;
}
}
private int findParent(int u) {
if (parent[u] == u) return u;
return parent[u] = findParent(parent[u]);
}
private boolean union(int u, int v) {
// System.out.println("uf "+u+" "+v);
int pu = findParent(u);
// System.out.println("uf2 "+pu+" "+v);
int pv = findParent(v);
// System.out.println("uf3 " + u + " " + pv);
if (pu == pv) return false;
if (size[pu] <= size[pv]) {
parent[pu] = pv;
size[pv] += size[pu];
} else {
parent[pv] = pu;
size[pu] += size[pv];
}
return true;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | b1d91e75d3aab577f61f387b2aa0fafe | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.sqrt;
import static java.lang.Math.pow;
import static java.lang.System.out;
import static java.lang.System.err;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static FastReader sc;
static long mod=((long)1e9)+7;
// static FastWriter out;
public static void main(String hi[]){
initializeIO();
sc=new FastReader();
int t=sc.nextInt();
// boolean[] seave=sieveOfEratosthenes((int)(1e7));
// int[] seave2=sieveOfEratosthenesInt((int)(1e4));
// int tp=1;
while(t--!=0){
int n=sc.nextInt();
long[][] arr=new long[n][2];
for(int i=0;i<n;i++){
arr[i][0]=sc.nextLong();
arr[i][1]=i+1;
}
Arrays.sort(arr,(a,b)->{
return (int)(a[0]-b[0]);
});
int size=n+(n-1);
long c=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if((arr[i][0]*arr[j][0])>size)break;
if((arr[i][0]*arr[j][0])==(arr[i][1]+arr[j][1]))c++;
}
}
out.println(c);
}
// System.out.println(String.format("%.9f", max));
}
static Boolean[][] dp;
static List<Integer> res=new ArrayList<>();
private static boolean solve(long[] arr,long w,long c,int i,List<Integer> li){
int n=arr.length;
long mid=(long)Math.ceil((1f*w)/(1f*2));
if(c>=(mid)&&c<=w){
res=new ArrayList<>(li);
return true;
}
if(i>=n||c>w)return false;
if(dp[i][(int)c]!=null)return dp[i][(int)c];
li.add(i+1);
boolean left=false;
if(c+arr[i]<=w)left=left|solve(arr,w,c+arr[i],i+1,li);
li.remove(li.size()-1);
boolean right=solve(arr,w,c,i+1,li);
return dp[i][(int)c]=left|right;
}
private static int sizeOfSubstring(int st,int e){
int s=e-st+1;
return (s*(s+1))/2;
}
private static int lessThen(long[] nums,long val){
int i=0,l=0,r=nums.length-1;
while(l<=r){
int mid=l+((r-l)/2);
if(nums[mid]<=val){
i=mid;
l=mid+1;
}else{
r=mid-1;
}
}
return i;
}
private static int lessThen(List<Long> nums,long val){
int i=0,l=0,r=nums.size()-1;
while(l<=r){
int mid=l+((r-l)/2);
if(nums.get(mid)<=val){
i=mid;
l=mid+1;
}else{
r=mid-1;
}
}
return i;
}
private static int greaterThen(long[] nums,long val){
int i=nums.length-1,l=0,r=nums.length-1;
while(l<=r){
int mid=l+((r-l)/2);
if(nums[mid]>=val){
i=mid;
r=mid-1;
}else{
l=mid+1;
}
}
return i;
}
private static long sumOfAp(long a,long n,long d){
long val=(n*( 2*a+((n-1)*d)));
return val/2;
}
//geometrics
private static double areaOftriangle(double x1,double y1,double x2,double y2,double x3,double y3){
double[] mid_point=midOfaLine(x1,y1,x2,y2);
// debug(Arrays.toString(mid_point)+" "+x1+" "+y1+" "+x2+" "+y2+" "+x3+" "+" "+y3);
double height=distanceBetweenPoints(mid_point[0],mid_point[1],x3,y3);
double wight=distanceBetweenPoints(x1,y1,x2,y2);
// debug(height+" "+wight);
return (height*wight)/2;
}
private static double distanceBetweenPoints(double x1,double y1,double x2,double y2){
double x=x2-x1;
double y=y2-y1;
return sqrt(Math.pow(x,2)+Math.pow(y,2));
}
private static double[] midOfaLine(double x1,double y1,double x2,double y2){
double[] mid=new double[2];
mid[0]=(x1+x2)/2;
mid[1]=(y1+y2)/2;
return mid;
}
private static long sumOfN(long n){
return (n*(n+1))/2;
}
private static long power(long x,long y,long p){
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0){
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
/* Function to calculate x raised to the power y in O(logn)*/
static long power(long x, long y){
long temp;
if( y == 0)
return 1l;
temp = power(x, y / 2);
if (y % 2 == 0)
return (temp*temp);
else
return (x*temp*temp);
}
private static StringBuilder reverseString(String s){
StringBuilder sb=new StringBuilder(s);
int l=0,r=sb.length()-1;
while(l<=r){
char ch=sb.charAt(l);
sb.setCharAt(l,sb.charAt(r));
sb.setCharAt(r,ch);
l++;
r--;
}
return sb;
}
private static void swap(long arr[],int i,int j){
long t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
private static void swap(List<Integer> li,int i,int j){
int t=li.get(i);
li.set(i,li.get(j));
li.set(j,t);
}
private static void swap(int arr[],int i,int j){
int t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
private static void swap(double arr[],int i,int j){
double t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
private static void swap(float arr[],int i,int j){
float t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static long lcm(long a, long b){
return (a / gcd(a, b)) * b;
}
private static String decimalToString(int x){
return Integer.toBinaryString(x);
}
private static boolean isPallindrome(String s){
int l=0,r=s.length()-1;
while(l<r){
if(s.charAt(l)!=s.charAt(r))return false;
l++;
r--;
}
return true;
}
private static StringBuilder removeLeadingZero(StringBuilder sb){
int i=0;
while(i<sb.length()&&sb.charAt(i)=='0')i++;
// debug("remove "+i);
if(i==sb.length())return new StringBuilder();
return new StringBuilder(sb.substring(i,sb.length()));
}
private static StringBuilder removeEndingZero(StringBuilder sb){
int i=sb.length()-1;
while(i>=0&&sb.charAt(i)=='0')i--;
// debug("remove "+i);
if(i<0)return new StringBuilder();
return new StringBuilder(sb.substring(0,i+1));
}
private static int stringToDecimal(String binaryString){
// debug(decimalToString(n<<1));
return Integer.parseInt(binaryString,2);
}
private static int stringToInt(String s){
return Integer.parseInt(s);
}
private static String toString(int val){
return String.valueOf(val);
}
private static void debug(int[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(long[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(List<int[]> arr){
for(int[] a:arr){
err.println(Arrays.toString(a));
}
}
private static void debug(float[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(double[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(boolean[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void print(String s){
out.println(s);
}
private static void debug(String s){
err.println(s);
}
private static int charToIntS(char c){
return ((((int)(c-'0'))%48));
}
private static void print(double s){
out.println(s);
}
private static void print(float s){
out.println(s);
}
private static void print(long s){
out.println(s);
}
private static void print(int s){
out.println(s);
}
private static void debug(double s){
err.println(s);
}
private static void debug(float s){
err.println(s);
}
private static void debug(long s){
err.println(s);
}
private static void debug(int s){
err.println(s);
}
private static boolean isPrime(int n){
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
//read graph
private static List<List<Integer>> readUndirectedGraph(int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<n;i++){
int x=sc.nextInt();
int y=sc.nextInt();
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readUndirectedGraph(int[][] intervals,int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<intervals.length;i++){
int x=intervals[i][0];
int y=intervals[i][1];
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int[][] intervals,int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<intervals.length;i++){
int x=intervals[i][0];
int y=intervals[i][1];
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<n;i++){
int x=sc.nextInt();
int y=sc.nextInt();
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
static String[] readStringArray(int n){
String[] arr=new String[n];
for(int i=0;i<n;i++){
arr[i]=sc.next();
}
return arr;
}
private static Map<Character,Integer> freq(String s){
Map<Character,Integer> map=new HashMap<>();
for(char c:s.toCharArray()){
map.put(c,map.getOrDefault(c,0)+1);
}
return map;
}
static boolean[] sieveOfEratosthenes(long n){
boolean prime[] = new boolean[(int)n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true){
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static int[] sieveOfEratosthenesInt(long n){
boolean prime[] = new boolean[(int)n + 1];
Set<Integer> li=new HashSet<>();
for (int i = 1; i <= n; i++){
prime[i] = true;
li.add(i);
}
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true){
for (int i = p * p; i <= n; i += p){
li.remove(i);
prime[i] = false;
}
}
}
int[] arr=new int[li.size()+1];
int i=0;
for(int x:li){
arr[i++]=x;
}
return arr;
}
public static long Kadens(List<Long> prices) {
long sofar=0;
long max_v=0;
for(int i=0;i<prices.size();i++){
sofar+=prices.get(i);
if (sofar<0) {
sofar=0;
}
max_v=Math.max(max_v,sofar);
}
return max_v;
}
public static int Kadens(int[] prices) {
int sofar=0;
int max_v=0;
for(int i=0;i<prices.length;i++){
sofar+=prices[i];
if (sofar<0) {
sofar=0;
}
max_v=Math.max(max_v,sofar);
}
return max_v;
}
static boolean isMemberAC(int a, int d, int x){
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
private static void sort(int[] arr){
int n=arr.length;
List<Integer> li=new ArrayList<>();
for(int x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(int[] arr){
int n=arr.length;
List<Integer> li=new ArrayList<>();
for(int x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sort(double[] arr){
int n=arr.length;
List<Double> li=new ArrayList<>();
for(double x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(double[] arr){
int n=arr.length;
List<Double> li=new ArrayList<>();
for(double x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(long[] arr){
int n=arr.length;
List<Long> li=new ArrayList<>();
for(long x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sort(long[] arr){
int n=arr.length;
List<Long> li=new ArrayList<>();
for(long x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static long sum(int[] arr){
long sum=0;
for(int x:arr){
sum+=x;
}
return sum;
}
private static long evenSumFibo(long n){
long l1=0,l2=2;
long sum=0;
while (l2<n) {
long l3=(4*l2)+l1;
sum+=l2;
if(l3>n)break;
l1=l2;
l2=l3;
}
return sum;
}
private static void initializeIO(){
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
System.setErr(new PrintStream(new FileOutputStream("error.txt")));
} catch (Exception e) {
// System.err.println(e.getMessage());
}
}
private static int maxOfArray(int[] arr){
int max=Integer.MIN_VALUE;
for(int x:arr){
max=Math.max(max,x);
}
return max;
}
private static long maxOfArray(long[] arr){
long max=Long.MIN_VALUE;
for(long x:arr){
max=Math.max(max,x);
}
return max;
}
private static int[][] readIntIntervals(int n,int m){
int[][] arr=new int[n][m];
for(int j=0;j<n;j++){
for(int i=0;i<m;i++){
arr[j][i]=sc.nextInt();
}
}
return arr;
}
private static long gcd(long a,long b){
if(b==0)return a;
return gcd(b,a%b);
}
private static int gcd(int a,int b){
if(b==0)return a;
return gcd(b,a%b);
}
private static int[] readIntArray(int n){
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
return arr;
}
private static double[] readDoubleArray(int n){
double[] arr=new double[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextDouble();
}
return arr;
}
private static long[] readLongArray(int n){
long[] arr=new long[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextLong();
}
return arr;
}
private static void print(int[] arr){
out.println(Arrays.toString(arr));
}
private static void print(long[] arr){
out.println(Arrays.toString(arr));
}
private static void print(String[] arr){
out.println(Arrays.toString(arr));
}
private static void print(double[] arr){
out.println(Arrays.toString(arr));
}
private static void debug(String[] arr){
err.println(Arrays.toString(arr));
}
private static void debug(Boolean[][] arr){
for(int i=0;i<arr.length;i++)err.println(Arrays.toString(arr[i]));
}
private static void debug(int[] arr){
err.println(Arrays.toString(arr));
}
private static void debug(long[] arr){
err.println(Arrays.toString(arr));
}
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while(st==null || !st.hasMoreTokens()){
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
static class Dsu {
int[] parent, size;
Dsu(int n) {
parent = new int[n + 1];
size = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
size[i] = 1;
}
}
private int findParent(int u) {
if (parent[u] == u) return u;
return parent[u] = findParent(parent[u]);
}
private boolean union(int u, int v) {
// System.out.println("uf "+u+" "+v);
int pu = findParent(u);
// System.out.println("uf2 "+pu+" "+v);
int pv = findParent(v);
// System.out.println("uf3 " + u + " " + pv);
if (pu == pv) return false;
if (size[pu] <= size[pv]) {
parent[pu] = pv;
size[pv] += size[pu];
} else {
parent[pv] = pu;
size[pu] += size[pv];
}
return true;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | d9112d8d373fcbd55d33b9e931c2e803 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.Math.sqrt;
import static java.lang.Math.pow;
import static java.lang.System.out;
import static java.lang.System.err;
import java.util.*;
import java.io.*;
import java.math.*;
public class Main
{
static FastReader sc;
static long mod=((long)1e9)+7;
// static FastWriter out;
public static void main(String hi[]){
initializeIO();
sc=new FastReader();
int t=sc.nextInt();
// boolean[] seave=sieveOfEratosthenes((int)(1e7));
// int[] seave2=sieveOfEratosthenesInt((int)(1e4));
// int tp=1;
while(t--!=0){
int n=sc.nextInt();
long[][] arr=new long[n][2];
for(int i=0;i<n;i++){
arr[i][0]=sc.nextLong();
arr[i][1]=i+1;
}
Arrays.sort(arr,(a,b)->{
return (int)(a[0]-b[0]);
});
int size=n+(n-1);
long c=0;
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if((arr[i][0]*arr[j][0])>size)break;
if((arr[i][0]*arr[j][0])==(arr[i][1]+arr[j][1]))c++;
}
}
out.println(c);
}
// System.out.println(String.format("%.9f", max));
}
static Boolean[][] dp;
static List<Integer> res=new ArrayList<>();
private static boolean solve(long[] arr,long w,long c,int i,List<Integer> li){
int n=arr.length;
long mid=(long)Math.ceil((1f*w)/(1f*2));
if(c>=(mid)&&c<=w){
res=new ArrayList<>(li);
return true;
}
if(i>=n||c>w)return false;
if(dp[i][(int)c]!=null)return dp[i][(int)c];
li.add(i+1);
boolean left=false;
if(c+arr[i]<=w)left=left|solve(arr,w,c+arr[i],i+1,li);
li.remove(li.size()-1);
boolean right=solve(arr,w,c,i+1,li);
return dp[i][(int)c]=left|right;
}
private static int sizeOfSubstring(int st,int e){
int s=e-st+1;
return (s*(s+1))/2;
}
private static int lessThen(long[] nums,long val){
int i=0,l=0,r=nums.length-1;
while(l<=r){
int mid=l+((r-l)/2);
if(nums[mid]<=val){
i=mid;
l=mid+1;
}else{
r=mid-1;
}
}
return i;
}
private static int lessThen(List<Long> nums,long val){
int i=0,l=0,r=nums.size()-1;
while(l<=r){
int mid=l+((r-l)/2);
if(nums.get(mid)<=val){
i=mid;
l=mid+1;
}else{
r=mid-1;
}
}
return i;
}
private static int greaterThen(long[] nums,long val){
int i=nums.length-1,l=0,r=nums.length-1;
while(l<=r){
int mid=l+((r-l)/2);
if(nums[mid]>=val){
i=mid;
r=mid-1;
}else{
l=mid+1;
}
}
return i;
}
private static long sumOfAp(long a,long n,long d){
long val=(n*( 2*a+((n-1)*d)));
return val/2;
}
//geometrics
private static double areaOftriangle(double x1,double y1,double x2,double y2,double x3,double y3){
double[] mid_point=midOfaLine(x1,y1,x2,y2);
// debug(Arrays.toString(mid_point)+" "+x1+" "+y1+" "+x2+" "+y2+" "+x3+" "+" "+y3);
double height=distanceBetweenPoints(mid_point[0],mid_point[1],x3,y3);
double wight=distanceBetweenPoints(x1,y1,x2,y2);
// debug(height+" "+wight);
return (height*wight)/2;
}
private static double distanceBetweenPoints(double x1,double y1,double x2,double y2){
double x=x2-x1;
double y=y2-y1;
return sqrt(Math.pow(x,2)+Math.pow(y,2));
}
private static double[] midOfaLine(double x1,double y1,double x2,double y2){
double[] mid=new double[2];
mid[0]=(x1+x2)/2;
mid[1]=(y1+y2)/2;
return mid;
}
private static long sumOfN(long n){
return (n*(n+1))/2;
}
private static long power(long x,long y,long p){
long res = 1; // Initialize result
x = x % p; // Update x if it is more than or
// equal to p
if (x == 0)
return 0; // In case x is divisible by p;
while (y > 0){
// If y is odd, multiply x with result
if ((y & 1) != 0)
res = (res * x) % p;
// y must be even now
y = y >> 1; // y = y/2
x = (x * x) % p;
}
return res;
}
/* Function to calculate x raised to the power y in O(logn)*/
static long power(long x, long y){
long temp;
if( y == 0)
return 1l;
temp = power(x, y / 2);
if (y % 2 == 0)
return (temp*temp);
else
return (x*temp*temp);
}
private static StringBuilder reverseString(String s){
StringBuilder sb=new StringBuilder(s);
int l=0,r=sb.length()-1;
while(l<=r){
char ch=sb.charAt(l);
sb.setCharAt(l,sb.charAt(r));
sb.setCharAt(r,ch);
l++;
r--;
}
return sb;
}
private static void swap(long arr[],int i,int j){
long t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
private static void swap(List<Integer> li,int i,int j){
int t=li.get(i);
li.set(i,li.get(j));
li.set(j,t);
}
private static void swap(int arr[],int i,int j){
int t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
private static void swap(double arr[],int i,int j){
double t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
private static void swap(float arr[],int i,int j){
float t=arr[i];
arr[i]=arr[j];
arr[j]=t;
}
static int lcm(int a, int b)
{
return (a / gcd(a, b)) * b;
}
static long lcm(long a, long b){
return (a / gcd(a, b)) * b;
}
private static String decimalToString(int x){
return Integer.toBinaryString(x);
}
private static boolean isPallindrome(String s){
int l=0,r=s.length()-1;
while(l<r){
if(s.charAt(l)!=s.charAt(r))return false;
l++;
r--;
}
return true;
}
private static StringBuilder removeLeadingZero(StringBuilder sb){
int i=0;
while(i<sb.length()&&sb.charAt(i)=='0')i++;
// debug("remove "+i);
if(i==sb.length())return new StringBuilder();
return new StringBuilder(sb.substring(i,sb.length()));
}
private static StringBuilder removeEndingZero(StringBuilder sb){
int i=sb.length()-1;
while(i>=0&&sb.charAt(i)=='0')i--;
// debug("remove "+i);
if(i<0)return new StringBuilder();
return new StringBuilder(sb.substring(0,i+1));
}
private static int stringToDecimal(String binaryString){
// debug(decimalToString(n<<1));
return Integer.parseInt(binaryString,2);
}
private static int stringToInt(String s){
return Integer.parseInt(s);
}
private static String toString(int val){
return String.valueOf(val);
}
private static void debug(int[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(long[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(List<int[]> arr){
for(int[] a:arr){
err.println(Arrays.toString(a));
}
}
private static void debug(float[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(double[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void debug(boolean[][] arr){
for(int i=0;i<arr.length;i++){
err.println(Arrays.toString(arr[i]));
}
}
private static void print(String s){
out.println(s);
}
private static void debug(String s){
err.println(s);
}
private static int charToIntS(char c){
return ((((int)(c-'0'))%48));
}
private static void print(double s){
out.println(s);
}
private static void print(float s){
out.println(s);
}
private static void print(long s){
out.println(s);
}
private static void print(int s){
out.println(s);
}
private static void debug(double s){
err.println(s);
}
private static void debug(float s){
err.println(s);
}
private static void debug(long s){
err.println(s);
}
private static void debug(int s){
err.println(s);
}
private static boolean isPrime(int n){
// Check if number is less than
// equal to 1
if (n <= 1)
return false;
// Check if number is 2
else if (n == 2)
return true;
// Check if n is a multiple of 2
else if (n % 2 == 0)
return false;
// If not, then just check the odds
for (int i = 3; i <= Math.sqrt(n); i += 2)
{
if (n % i == 0)
return false;
}
return true;
}
//read graph
private static List<List<Integer>> readUndirectedGraph(int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<n;i++){
int x=sc.nextInt();
int y=sc.nextInt();
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readUndirectedGraph(int[][] intervals,int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<intervals.length;i++){
int x=intervals[i][0];
int y=intervals[i][1];
graph.get(x).add(y);
graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int[][] intervals,int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<intervals.length;i++){
int x=intervals[i][0];
int y=intervals[i][1];
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
private static List<List<Integer>> readDirectedGraph(int n){
List<List<Integer>> graph=new ArrayList<>();
for(int i=0;i<=n;i++){
graph.add(new ArrayList<>());
}
for(int i=0;i<n;i++){
int x=sc.nextInt();
int y=sc.nextInt();
graph.get(x).add(y);
// graph.get(y).add(x);
}
return graph;
}
static String[] readStringArray(int n){
String[] arr=new String[n];
for(int i=0;i<n;i++){
arr[i]=sc.next();
}
return arr;
}
private static Map<Character,Integer> freq(String s){
Map<Character,Integer> map=new HashMap<>();
for(char c:s.toCharArray()){
map.put(c,map.getOrDefault(c,0)+1);
}
return map;
}
static boolean[] sieveOfEratosthenes(long n){
boolean prime[] = new boolean[(int)n + 1];
for (int i = 2; i <= n; i++)
prime[i] = true;
for (int p = 2; p * p <= n; p++)
{
// If prime[p] is not changed, then it is a
// prime
if (prime[p] == true){
// Update all multiples of p
for (int i = p * p; i <= n; i += p)
prime[i] = false;
}
}
return prime;
}
static int[] sieveOfEratosthenesInt(long n){
boolean prime[] = new boolean[(int)n + 1];
Set<Integer> li=new HashSet<>();
for (int i = 1; i <= n; i++){
prime[i] = true;
li.add(i);
}
for (int p = 2; p * p <= n; p++)
{
if (prime[p] == true){
for (int i = p * p; i <= n; i += p){
li.remove(i);
prime[i] = false;
}
}
}
int[] arr=new int[li.size()+1];
int i=0;
for(int x:li){
arr[i++]=x;
}
return arr;
}
public static long Kadens(List<Long> prices) {
long sofar=0;
long max_v=0;
for(int i=0;i<prices.size();i++){
sofar+=prices.get(i);
if (sofar<0) {
sofar=0;
}
max_v=Math.max(max_v,sofar);
}
return max_v;
}
public static int Kadens(int[] prices) {
int sofar=0;
int max_v=0;
for(int i=0;i<prices.length;i++){
sofar+=prices[i];
if (sofar<0) {
sofar=0;
}
max_v=Math.max(max_v,sofar);
}
return max_v;
}
static boolean isMemberAC(int a, int d, int x){
// If difference is 0, then x must
// be same as a.
if (d == 0)
return (x == a);
// Else difference between x and a
// must be divisible by d.
return ((x - a) % d == 0 && (x - a) / d >= 0);
}
private static void sort(int[] arr){
int n=arr.length;
List<Integer> li=new ArrayList<>();
for(int x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(int[] arr){
int n=arr.length;
List<Integer> li=new ArrayList<>();
for(int x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sort(double[] arr){
int n=arr.length;
List<Double> li=new ArrayList<>();
for(double x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(double[] arr){
int n=arr.length;
List<Double> li=new ArrayList<>();
for(double x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sortReverse(long[] arr){
int n=arr.length;
List<Long> li=new ArrayList<>();
for(long x:arr){
li.add(x);
}
Collections.sort(li,Collections.reverseOrder());
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static void sort(long[] arr){
int n=arr.length;
List<Long> li=new ArrayList<>();
for(long x:arr){
li.add(x);
}
Collections.sort(li);
for (int i=0;i<n;i++) {
arr[i]=li.get(i);
}
}
private static long sum(int[] arr){
long sum=0;
for(int x:arr){
sum+=x;
}
return sum;
}
private static long evenSumFibo(long n){
long l1=0,l2=2;
long sum=0;
while (l2<n) {
long l3=(4*l2)+l1;
sum+=l2;
if(l3>n)break;
l1=l2;
l2=l3;
}
return sum;
}
private static void initializeIO(){
try {
System.setIn(new FileInputStream("input.txt"));
System.setOut(new PrintStream(new FileOutputStream("output.txt")));
System.setErr(new PrintStream(new FileOutputStream("error.txt")));
} catch (Exception e) {
// System.err.println(e.getMessage());
}
}
private static int maxOfArray(int[] arr){
int max=Integer.MIN_VALUE;
for(int x:arr){
max=Math.max(max,x);
}
return max;
}
private static long maxOfArray(long[] arr){
long max=Long.MIN_VALUE;
for(long x:arr){
max=Math.max(max,x);
}
return max;
}
private static int[][] readIntIntervals(int n,int m){
int[][] arr=new int[n][m];
for(int j=0;j<n;j++){
for(int i=0;i<m;i++){
arr[j][i]=sc.nextInt();
}
}
return arr;
}
private static long gcd(long a,long b){
if(b==0)return a;
return gcd(b,a%b);
}
private static int gcd(int a,int b){
if(b==0)return a;
return gcd(b,a%b);
}
private static int[] readIntArray(int n){
int[] arr=new int[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextInt();
}
return arr;
}
private static double[] readDoubleArray(int n){
double[] arr=new double[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextDouble();
}
return arr;
}
private static long[] readLongArray(int n){
long[] arr=new long[n];
for(int i=0;i<n;i++){
arr[i]=sc.nextLong();
}
return arr;
}
private static void print(int[] arr){
out.println(Arrays.toString(arr));
}
private static void print(long[] arr){
out.println(Arrays.toString(arr));
}
private static void print(String[] arr){
out.println(Arrays.toString(arr));
}
private static void print(double[] arr){
out.println(Arrays.toString(arr));
}
private static void debug(String[] arr){
err.println(Arrays.toString(arr));
}
private static void debug(Boolean[][] arr){
for(int i=0;i<arr.length;i++)err.println(Arrays.toString(arr[i]));
}
private static void debug(int[] arr){
err.println(Arrays.toString(arr));
}
private static void debug(long[] arr){
err.println(Arrays.toString(arr));
}
static class FastReader{
BufferedReader br;
StringTokenizer st;
public FastReader(){
br=new BufferedReader(new InputStreamReader(System.in));
}
String next(){
while(st==null || !st.hasMoreTokens()){
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt(){
return Integer.parseInt(next());
}
long nextLong(){
return Long.parseLong(next());
}
double nextDouble(){
return Double.parseDouble(next());
}
String nextLine(){
String str="";
try {
str=br.readLine().trim();
} catch (Exception e) {
e.printStackTrace();
}
return str;
}
}
static class FastWriter {
private final BufferedWriter bw;
public FastWriter() {
this.bw = new BufferedWriter(new OutputStreamWriter(System.out));
}
public void print(Object object) throws IOException {
bw.append("" + object);
}
public void println(Object object) throws IOException {
print(object);
bw.append("\n");
}
public void close() throws IOException {
bw.close();
}
}
static class Dsu {
int[] parent, size;
Dsu(int n) {
parent = new int[n + 1];
size = new int[n + 1];
for (int i = 0; i <= n; i++) {
parent[i] = i;
size[i] = 1;
}
}
private int findParent(int u) {
if (parent[u] == u) return u;
return parent[u] = findParent(parent[u]);
}
private boolean union(int u, int v) {
// System.out.println("uf "+u+" "+v);
int pu = findParent(u);
// System.out.println("uf2 "+pu+" "+v);
int pv = findParent(v);
// System.out.println("uf3 " + u + " " + pv);
if (pu == pv) return false;
if (size[pu] <= size[pv]) {
parent[pu] = pv;
size[pv] += size[pu];
} else {
parent[pv] = pu;
size[pu] += size[pv];
}
return true;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 702f7ccc13b0b4f6e59f5891946fe5d1 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class Pleasant_Pairs {
public static void main(String[] args) {
// TODO Auto-generated method stub
Scanner sc = new Scanner(System.in);
StringBuilder sb = new StringBuilder();
int t = sc.nextInt();
while(t-->0) {
int n = sc.nextInt();
int arr[] = new int[n+1];
for(int i=1;i<=n;i++)
arr[i] = sc.nextInt();
int count=0;
for(int i=1;i<n;i++) {
for(int j = arr[i]-i;j<=n;j+= arr[i])
if(j>1)
if(i<j&&((long)arr[i]*arr[j])==(i+j))
count++;
}
sb.append(count+"\n");
}
System.out.println(sb);
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 3ec7763338940f471e0f6965cd28ccd0 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class Pattern {
public static void main(String[] args) {
Scanner scan = new Scanner(System.in);
int t = scan.nextInt();
while(t-->0) {
int n = scan.nextInt();
int a[] = new int[n+1];
for(int i =1;i<=n;i++)
a[i] = scan.nextInt();
int count =0;
for(long i =1;i<=n;i++) {
for(long j = a[(int)i]-i;j<=n;j+=a[(int)i]) {
if(j>0)
if(j>i&&1l*a[(int)i]*a[(int)j]==0l+i+j) {
count++;
}
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 81c5889509bfb445b74782fae425cb08 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
public class cp
{
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main(String[] args) throws IOException
{
try{
//Your Solve
Reader s = new Reader();
int t = s.nextInt();
for(int p = 0;p < t;p++){
int n = s.nextInt();
int a[] = new int[n+1];
for(int i = 1;i < n+1;i++) {
a[i] = s.nextInt();
}
int count = 0;
for(int i = 1;i < n;i++) {
for(int sum = 2*i+1 + (((2*i+1)%a[i] == 0)?0:a[i]-((2*i+1)%a[i]));sum-i < n+1;sum += a[i]) {
// System.out.println(i + " " + (sum-i));
if(a[sum-i] == sum/a[i] && sum % a[sum-i] == 0) {
count++;
}
}
}
System.out.println(count);
}
}catch(Exception e){
return;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | d77c5d6fef882f55f4a1dc5796509a53 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.*;
import java.util.*;
public class cp
{
static class Reader {
final private int BUFFER_SIZE = 1 << 16;
private DataInputStream din;
private byte[] buffer;
private int bufferPointer, bytesRead;
public Reader()
{
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public Reader(String file_name) throws IOException
{
din = new DataInputStream(
new FileInputStream(file_name));
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException
{
byte[] buf = new byte[64]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
if (cnt != 0) {
break;
}
else {
continue;
}
}
buf[cnt++] = (byte)c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException
{
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public long nextLong() throws IOException
{
long ret = 0;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg)
return -ret;
return ret;
}
public double nextDouble() throws IOException
{
double ret = 0, div = 1;
byte c = read();
while (c <= ' ')
c = read();
boolean neg = (c == '-');
if (neg)
c = read();
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg)
return -ret;
return ret;
}
private void fillBuffer() throws IOException
{
bytesRead = din.read(buffer, bufferPointer = 0,
BUFFER_SIZE);
if (bytesRead == -1)
buffer[0] = -1;
}
private byte read() throws IOException
{
if (bufferPointer == bytesRead)
fillBuffer();
return buffer[bufferPointer++];
}
public void close() throws IOException
{
if (din == null)
return;
din.close();
}
}
public static void main(String[] args) throws IOException
{
try{
//Your Solve
Reader s = new Reader();
int t = s.nextInt();
for(int p = 0;p < t;p++){
int n = s.nextInt();
int a[] = new int[n+1];
for(int i = 1;i < n+1;i++) {
a[i] = s.nextInt();
}
int count = 0;
for(int i = 1;i < n;i++) {
if(a[i] == 1) {
for(int j = i+1;j < n+1;j++) {
if(a[j] == i+j) {
count++;
}
}
}else {
for(int sum = 2*i+1 + (((2*i+1)%a[i] == 0)?0:a[i]-((2*i+1)%a[i]));sum-i < n+1;sum += a[i]) {
// System.out.println(i + " " + (sum-i));
if(a[sum-i] == sum/a[i] && sum % a[sum-i] == 0) {
count++;
}
}
}
}
System.out.println(count);
}
}catch(Exception e){
return;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 360879d5d4a28f56b85b84edfef60b79 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.DataInputStream;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.Comparator;
public class Main {
private static void run() throws IOException {
int n = in.nextInt();
Node[] a = new Node[n];
for (int i = 0; i < n; i++) {
a[i] = new Node(i + 1, in.nextInt());
}
Arrays.sort(a, Comparator.comparingLong(o -> o.value));
long ans = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < i; j++) {
long now = a[i].value * a[j].value;
if (now == a[i].index + a[j].index) {
ans++;
}
if (now > 2L * n) {
break;
}
}
}
out.println(ans);
}
static class Node {
long index, value;
public Node(long index, long value) {
this.value = value;
this.index = index;
}
}
public static void main(String[] args) throws IOException {
in = new Reader();
out = new PrintWriter(new OutputStreamWriter(System.out));
int t = in.nextInt();
for (int i = 0; i < t; i++) {
run();
}
out.flush();
in.close();
out.close();
}
private static int gcd(int a, int b) {
if (a == 0 || b == 0)
return 0;
while (b != 0) {
int tmp;
tmp = a % b;
a = b;
b = tmp;
}
return a;
}
static final long mod = 1000000007;
static long pow_mod(long a, long b) {
long result = 1;
while (b != 0) {
if ((b & 1) != 0) result = (result * a) % mod;
a = (a * a) % mod;
b >>= 1;
}
return result;
}
private static long multiplied_mod(long... longs) {
long ans = 1;
for (long now : longs) {
ans = (ans * now) % mod;
}
return ans;
}
@SuppressWarnings("FieldCanBeLocal")
private static Reader in;
private static PrintWriter out;
private static int[] read_int_array(int len) throws IOException {
int[] a = new int[len];
for (int i = 0; i < len; i++) {
a[i] = in.nextInt();
}
return a;
}
private static long[] read_long_array(int len) throws IOException {
long[] a = new long[len];
for (int i = 0; i < len; i++) {
a[i] = in.nextLong();
}
return a;
}
private static void print_array(int[] array) {
for (int now : array) {
out.print(now);
out.print(' ');
}
out.println();
}
private static void print_array(long[] array) {
for (long now : array) {
out.print(now);
out.print(' ');
}
out.println();
}
static class Reader {
private static final int BUFFER_SIZE = 1 << 16;
private final DataInputStream din;
private final byte[] buffer;
private int bufferPointer, bytesRead;
Reader() {
din = new DataInputStream(System.in);
buffer = new byte[BUFFER_SIZE];
bufferPointer = bytesRead = 0;
}
public String readLine() throws IOException {
final byte[] buf = new byte[1024]; // line length
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\n') {
break;
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextSign() throws IOException {
byte c = read();
while ('+' != c && '-' != c) {
c = read();
}
return '+' == c ? 0 : 1;
}
private static boolean isSpaceChar(int c) {
return !(c >= 33 && c <= 126);
}
public int skip() throws IOException {
int b;
// noinspection ALL
while ((b = read()) != -1 && isSpaceChar(b)) {
;
}
return b;
}
public char nc() throws IOException {
return (char) skip();
}
public String next() throws IOException {
int b = skip();
final StringBuilder sb = new StringBuilder();
while (!isSpaceChar(b)) { // when nextLine, (isSpaceChar(b) && b != ' ')
sb.appendCodePoint(b);
b = read();
}
return sb.toString();
}
public int nextInt() throws IOException {
int ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
final boolean neg = c == '-';
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
} while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public long nextLong() throws IOException {
long ret = 0;
byte c = read();
while (c <= ' ') {
c = read();
}
final boolean neg = c == '-';
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (neg) {
return -ret;
}
return ret;
}
public double nextDouble() throws IOException {
double ret = 0, div = 1;
byte c = read();
while (c <= ' ') {
c = read();
}
final boolean neg = c == '-';
if (neg) {
c = read();
}
do {
ret = ret * 10 + c - '0';
}
while ((c = read()) >= '0' && c <= '9');
if (c == '.') {
while ((c = read()) >= '0' && c <= '9') {
ret += (c - '0') / (div *= 10);
}
}
if (neg) {
return -ret;
}
return ret;
}
private void fillBuffer() throws IOException {
bytesRead = din.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) {
buffer[0] = -1;
}
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) {
fillBuffer();
}
return buffer[bufferPointer++];
}
public void close() throws IOException {
din.close();
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 19bc02a37810deeeeb738f621d547c59 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Main{
public static long nPairs(int arr[],int n,HashMap<Integer,Integer> hmap){
long count=0;
int max=2*n;
for(int i=0;i<n;i++){
for(int j=1;j*arr[i]<=max;j++){
if(!hmap.containsKey(j))
continue;
if(hmap.get(j)>i+1 && j*arr[i]==hmap.get(j)+hmap.get(arr[i]))
count++;
}
}
return count;
}
public static void main (String[] args) throws java.lang.Exception
{
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
int t=Integer.parseInt(br.readLine());
while(t-->0){
int n=Integer.parseInt(br.readLine());
StringTokenizer st=new StringTokenizer(br.readLine());
int arr[]=new int[n];
HashMap<Integer,Integer> hmap=new HashMap<Integer,Integer>();
for(int i=0;i<n;i++){
arr[i]=Integer.parseInt(st.nextToken());
hmap.put(arr[i],i+1);
//System.out.print(arr[i]+" ");
}
//System.out.println("");
System.out.println(nPairs(arr,n,hmap));
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 242ab23ca6e4832206a0793bb2726cc6 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | /*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
public class GFG {
public static void main (String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-- > 0){
int n = scn.nextInt();
int arr[] = new int[n+1];
for(int i=1; i<=n; i++){
arr[i] = scn.nextInt();
}
int count = 0;
for(int i=1; i<=n; i++){
/* arr[i] * arr[j] = i + j
arr[i] * [1, 2, 3....] = i + ([1, 2, 3...] * arr[i] - i)
Example: at i = 1, there is 3.
so we will check at j = 2, arr[j] == 1 as (3 * 1 = i + j),
then check at j = 5, arr[j] == 2 as (3 * 2 = i + j),
above statement implies that for a particular index->i, we have to search for those indexes->j, where the sum of (i+j) is a multiple of value at 'i'
Example: 3 1 5 9 2
for i=1, arr[i]=3
A. 3 * arr[j] = 1 + (3-1), here j becomes 2
=> 3 * 1 = 1 + 2
B. 3 * arr[j] = 1` + (2*3 - 1), here j becomes 5
=> 3 * 2 = 1 + 5
for i=2, arr[i]=1
A. 1 * arr[j] = 2 + (1-2), here j becomes 1
=> 1 * ? != 2 - 1
B. 1 * arr[j] = 2 + (2*1 - 2), here, j becomes 0
=> 1 * ? != 2 +- 0
WON'T COUNT TILL J>I
C. 1 * arr[j] = 2 + (4*1 - 2), here, j becomes 5
=> 1 * 5? != 2 + 3
*/
int si = arr[i] - i;
int val = 1;
while(si<=n){
// second idx should be greater than first idx
if(i < si && val == arr[si]){
count++;
}
si += arr[i];
val++;
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | ab5fa20face971e5b205a2528a3c8c6b | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | /*package whatever //do not write package name here */
import java.io.*;
import java.util.*;
public class GFG {
public static void main (String[] args) {
Scanner scn = new Scanner(System.in);
int t = scn.nextInt();
while(t-- > 0){
int n = scn.nextInt();
int arr[] = new int[n+1];
for(int i=1; i<=n; i++){
arr[i] = scn.nextInt();
}
int count = 0;
for(int i=1; i<=n; i++){
int num = arr[i];
int si = num - i;
int val = 1;
while(si<=n){
if(i < si && val == arr[si]){
count++;
}
si += arr[i];
val++;
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 51a5c38207b1f6c86d40a74be8a07919 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.Scanner;
public class B {
public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int t = s.nextInt();
while(t-- > 0) {
int n = s.nextInt();
int[] a = new int[n + 1];
for(int i = 1; i <= n; i++) a[i] = s.nextInt();
long res = 0;
for(int i = 1; i < n; i++) {
int k = 1;
while(k * a[i] - i <= n) {
if(k * a[i] - i > i && a[k * a[i] - i] == k) res++;
k++;
}
}
System.out.println(res);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 5efd8e585920f3d3a75369e413b771a7 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigInteger;
import java.util.*;
import static java.lang.System.out;
public class Solution {
static MyScanner str = new MyScanner();
// static Scanner str = new Scanner(System.in);
// static Set<String> set;
int mod = (int)1e9 + 7;
public static void main(String[] args) throws IOException {
long T = l();
while (T-- > 0) {
solve();
}
}
static void solve() throws IOException {
int n = i();
int a[] = new int[n + 1];
for (int i = 1; i <= n; i++) {
a[i] = i();
}
int ans = 0;
for (int i = 1; i <= n; i++) {
for (int j = a[i] - i; j <= n; j += a[i]) {
if (j >= 0) {
if ((long) a[i] * a[j] == i + j && i < j) {
ans++;
}
}
}
}
out.println(ans);
}
public static int i() throws IOException {
return str.nextInt();
}
public static long l() throws IOException {
return str.nextLong();
}
public static double d() throws IOException {
return str.nextDouble();
}
public static class MyScanner {
BufferedReader br;
StringTokenizer st;
public MyScanner() {
br = new BufferedReader(new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 6adcc7d728ba49a48b74dd126a67c1fd | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class Main {
public static void main(String[] args) throws java.lang.Exception{
Scanner sc=new Scanner(System.in);
int t=sc.nextInt();
while(t-->0) {
int n=sc.nextInt();
int a[]=new int[n+1];
for(int i=1;i<=n;i++) {
a[i]=sc.nextInt();
}
long c=0;
for(int i=1;i<n;i++) {
for(int j=a[i]-i;j<=n;j+=a[i]) {
if(j>=1)
if(((long)a[i]*a[j]==i+j)&&(i<j))
c++;
}
}
System.out.println(c);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 2b0657d649fc7ef831398d591416478d | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
// Working program with FastReader
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class B1541 {
public static void main(String[] args) {
FastReader fs = new FastReader();
int t = fs.nextInt();
StringBuilder sb = new StringBuilder();
while (t-- > 0) {
int n = fs.nextInt();
int[] arr = fs.readArray(n);
Map<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < n; ++i) {
map.put(arr[i], i + 1);
}
int min = 3;
int max = 2 * n;
int count = 0;
for (int i = min; i <= max; i++) {
int c = getCount(i, map);
count += c;
}
sb.append(count).append("\n");
}
System.out.println(sb);
}
private static int getCount(int min, Map<Integer, Integer> map) {
int c = 0;
for (int i = 1; i * i <= min; ++i) {
if (min % i == 0) {
int second = min / i;
if (map.containsKey(i) && map.containsKey(second)) {
int i1 = map.get(i);
int i2 = map.get(second);
if (i1 != i2 && i1 + i2 == min) {
c++;
}
}
}
}
return c;
}
private static boolean isPrime(long num) {
if (num == 2L) {
return true;
}
for (long i = 2; i * i <= num; i++) {
if (num % i == 0) {
return false;
}
}
return true;
}
private static long getGcd(long a, long b) {
if (b == 0) return a;
return getGcd(b, a % b);
}
private static long binaryExponentiation(long a, long b) {
long res = 1;
while (b > 0) {
if ((b & 1) == 1) res = res * a;
a = a * a;
b >>= 1;
}
return res;
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(next());
return arr;
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 2ffeee0302f0344d0a2efa989e0671f6 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
// Working program with FastReader
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.*;
public class PleasantPairs {
public static void main(String[] args) {
FastReader fs = new FastReader();
int t = fs.nextInt();
StringBuilder sb = new StringBuilder();
while (t-- > 0) {
int n = fs.nextInt();
long count = 0;
HashMap<Integer, Integer> map = new HashMap<>();
for (int i = 0; i < n; i++) map.put(fs.nextInt(), i + 1);
for (int i = 1; i <= 2 * n; i++) {
for (int j = i + 1; j <= 2 * n; j++) {
if ((long) i * j >= 2L * n) break;
if (map.containsKey(i) && map.containsKey(j)) {
if (((long) i * j) == (map.get(i) + map.get(j))) count++;
}
}
}
sb.append(count).append("\n");
}
System.out.println(sb);
}
private static int getGcd(int a, int b) {
if (b == 0) return a;
return getGcd(b, a % b);
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
int[] readArray(int n) {
int[] arr = new int[n];
for (int i = 0; i < n; i++) arr[i] = Integer.parseInt(next());
return arr;
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 198387ea007c66bc33b441ea99e9bd22 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.io.BufferedReader;
import java.lang.*;
import java.io.FileReader;
import java.io.IOException;
import java.math.BigInteger;
import java.io.IOException;
import java.io.InputStreamReader;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.math.MathContext;
import java.util.*;
import java.util.LinkedList;
public class Pleasent_Pair {
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader() {
br = new BufferedReader(new
InputStreamReader(System.in));
}
String next() {
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
long nextLong() {
return Long.parseLong(next());
}
double nextDouble() {
return Double.parseDouble(next());
}
String nextLine() {
String str = "";
try {
str = br.readLine();
} catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
static void sort(long[] arr) {
ArrayList<Long> a = new ArrayList<>();
for (long i : arr) {
a.add(i);
}
Collections.sort(a);
for (int i = 0; i < a.size(); i++) {
arr[i] = a.get(i);
}
}
static int get(long[] arr, int tar){
int left = 1;
int right = tar-1;
int ans = 0;
// System.out.println(tar);
while(left < right){
if(left-1 >= 0 &&left -1 < arr.length && right-1 >=0 && right-1< arr.length)
if(arr[left-1]*arr[right-1]== tar)ans++;
left++;
right--;
}
return ans;
}
public static void main(String[] args) throws Exception {
FastReader sc = new FastReader();
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
int[] arr = new int[n+1];
for (int i = 1; i <= n; i++) {
arr[i] = sc.nextInt();
}
long ans = 0;
for(int i = 1; i <n; i++){
for(int j = arr[i]-i; j <=n; j+=arr[i]){
if(j>=0)
if((long)arr[i]*arr[j] ==(long)i+j && i < j)ans++;
}
}
System.out.println(ans);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 4927d083864f099f8b420b44b7f243a9 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class B_Pleasant_Pairs {
public static void main(String[] args) throws java.lang.Exception{
Scanner sc=new Scanner(System.in);
int T=sc.nextInt();
while(T-->0) {
int n=sc.nextInt();
int a[]=new int[n+1];
for(int i=1;i<=n;i++) {
a[i]=sc.nextInt();
}
long c=0;
for(int i=1;i<n;i++) {
for(int j=a[i]-i;j<=n;j+=a[i]) {
if(j>=1)
if(((long)a[i]*a[j]==i+j)&&(i<j))
c++;
}
}
System.out.println(c);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 709a17b7df08f5a1e1bbffb66b15447c | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes |
import java.util.*;
public class A {
static class Pair {
long x;
long y;
public Pair(long x, long y) {
this.x = x;
this.y = y;
}
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
StringBuilder str = new StringBuilder();
int t = sc.nextInt();
for (int xx = 0; xx < t; xx++) {
int n = sc.nextInt();
Pair[] arr = new Pair[n];
for(int i=0;i<n;i++)
{
arr[i] = new Pair(sc.nextLong(),i+1);
}
Arrays.sort(arr ,(p1,p2)-> {
return (int)p1.x - (int)p2.x;
});
int c =0;
long l =1;
for(int i=0;i<n;i++)
{
for(int j=i+1;j<n;j++)
{
l = arr[i].x*arr[j].x;
if(l==arr[i].y+arr[j].y)
c++;
if(l>2*n)
break;
}
}
str.append(c+"\n");
}
System.out.print(str);
sc.close();
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 5b95648fc0ad92df69b2d9072fc7b3b4 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
//----------- StringBuilder for faster output------------------------------
static StringBuilder out = new StringBuilder();
public static void main(String[] args) {
FastScanner fs=new FastScanner();
/****** CODE STARTS HERE *****/
int t = fs.nextInt();
while(t-->0) {
int n = fs.nextInt();
int[] a = new int [2*n+1];
Arrays.fill(a, -1);
for(int i=1; i<=n; i++)a[fs.nextInt()]=i;
int cnt = 0;
for(int i=3; i<2*n; i++)
{
for(int j=1; j*j<i; j++)
{
if(i%j==0 && a[j]!=-1 && a[i/j]!= -1 && i==a[j]+a[i/j])cnt++;
}
}
out.append(cnt+"\n");
}
System.out.print(out);
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//----------- FastScanner class for faster input---------------------------
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 6ec60db79c13ed9b7ab83f9a4c4209df | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
//----------- StringBuilder for faster output------------------------------
static StringBuilder out = new StringBuilder();
public static void main(String[] args) {
FastScanner fs=new FastScanner();
/****** CODE STARTS HERE *****/
int t = fs.nextInt();
while(t-->0) {
int n = fs.nextInt();
int[] a = new int [2*n+1];
Arrays.fill(a, -1);
for(int i=1; i<=n; i++)a[fs.nextInt()]=i;
int cnt = 0;
for(int i=3; i<2*n; i++)
{
for(int j=1; j*j<=i; j++)
{
if(i%j==0 && j*j!=i && a[j]!=-1 && a[i/j]!= -1 && i==a[j]+a[i/j])cnt++;
}
}
out.append(cnt+"\n");
}
System.out.print(out);
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//----------- FastScanner class for faster input---------------------------
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | b696961e015df0bb4e197b9ea7ebfbfd | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.*;
import java.util.*;
public class Main {
//----------- StringBuilder for faster output------------------------------
static StringBuilder out = new StringBuilder();
public static void main(String[] args) {
FastScanner fs=new FastScanner();
/****** CODE STARTS HERE *****/
int t = fs.nextInt();
while(t-->0) {
int n = fs.nextInt();
int[] a = fs.readArray(n);
int cnt = 0;
for(int i=1; i<=n; i++)
{
for(int j=a[i-1]-i; j<=n; j+=a[i-1])
{
if(j>0 && i<j && (long)a[i-1]*a[j-1]==(i+j))cnt++;
}
}
out.append(cnt+"\n");
}
System.out.print(out);
}
static void sort(int[] a) {
ArrayList<Integer> l=new ArrayList<>();
for (int i:a) l.add(i);
Collections.sort(l);
for (int i=0; i<a.length; i++) a[i]=l.get(i);
}
//----------- FastScanner class for faster input---------------------------
static class FastScanner {
BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
StringTokenizer st=new StringTokenizer("");
String next() {
while (!st.hasMoreTokens())
try {
st=new StringTokenizer(br.readLine());
} catch (IOException e) {
e.printStackTrace();
}
return st.nextToken();
}
int nextInt() {
return Integer.parseInt(next());
}
int[] readArray(int n) {
int[] a=new int[n];
for (int i=0; i<n; i++) a[i]=nextInt();
return a;
}
long nextLong() {
return Long.parseLong(next());
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | eeea6ef0dee85f4c07dbe4f755ee14d9 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
public class PleasantPairs {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int t = sc.nextInt();
while (t-- > 0) {
int n = sc.nextInt();
long[] a = new long[n + 1];
for (int i = 1; i < n + 1; i++) {
a[i] = sc.nextInt();
}
long count = 0;
for (int i = 1; i <= n; i++) {
for (int j = (int) a[i] - i; j <= n; j += a[i]) {
if (j > 0) {
if (a[i] * a[j] == i + j && j > i) {
count++;
}
}
}
}
System.out.println(count);
}
}
} | Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | f22cccdeb7bd025883997bd0906f9767 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class B{
public static void main(String[] args){
FastReader sc = new FastReader();
int t=sc.nextInt();
while(t-->0){
int n=sc.nextInt();
int[]a=new int [n+1];
for(int i=1;i<=n;i++) {
a[i]=sc.nextInt();
}
int ans=0;
for(int i=1;i<=n;i++) {
for(int j=a[i]-i;j<=n;j+=a[i]) {
if(j>0 &&(long)a[i]*a[j]==(long)i+j && i<j)ans++;
}
}
System.out.println(ans);
}
}
static class ind {
int x;int y;
ind(int x,int y){
this.x=x;
this.y=y;
}
}
static class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 2b068635d8bcb635095567910f51252e | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.util.*;
import java.io.*;
public class Pleasant_Pairs {
public static void main(String[] args) throws java.lang.Exception {
BufferedReader input = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(input.readLine());
for(int i=0;i<t;i++){
int n = Integer.parseInt(input.readLine());
String[] s = input.readLine().split(" ");
Long[][] nums = new Long[n][2];
long count = 0;
for(int j=0;j<n;j++){
nums[j][0] = Long.parseLong(s[j]);
nums[j][1] = (long)(j+1);
}
sortbyColumn(nums, 0);
for(int k=0;k<n;k++){
for(int l=k+1;l<n;l++){
long val = nums[k][0]*nums[l][0];
if(val>2*n){
break;
}
if(val==nums[k][1]+nums[l][1]){
count++;
}
}
}
System.out.println(count);
}
}
public static void sortbyColumn(Long arr[][], int col)
{
Arrays.sort(arr, new Comparator<Long[]>() {
@Override
public int compare(final Long[] entry1,
final Long[] entry2) {
if (entry1[col] > entry2[col])
return 1;
else
return -1;
}
});
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | d181ab1087f7e936d1484c7a5f9436b3 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.util.TreeSet;
public class Main {
static boolean hasMultipleTestCases = true;
static final String newLine = "\n";
public static void main(String[] args) {
InputReader in = new InputReader(System.in);
PrintWriter out = new PrintWriter(System.out);
StringBuilder sb = new StringBuilder();
int testCases = 1;
if (hasMultipleTestCases) {
testCases = in.nextInt();
}
while (testCases-- > 0) {
int n = in.nextInt();
int[] a = new int[n + 1];
for (int i = 1; i <= n; i++) a[i] = in.nextInt();
int count = 0;
for (int i = 1; i <= n; i++) {
for (int j = a[i] - i; j <= n; j += a[i]) {
if (j > i) {
if ((long)i + j == (long)a[i] * a[j]) count++;
}
}
}
sb.append(count).append(newLine);
}
out.print(sb);
out.flush();
}
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1) throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0) return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9') throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9') throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c)) c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
boolean isSpaceChar(int ch);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | ba76d0153cf0c3b7987c698fd570fbf5 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.InputMismatchException;
import java.util.TreeSet;
public class Main {
static boolean hasMultipleTestCases = true;
static final String newLine = "\n";
public static void main(String[] args) {
InputReader in = new InputReader(System.in);
PrintWriter out = new PrintWriter(System.out);
StringBuilder sb = new StringBuilder();
int testCases = 1;
if (hasMultipleTestCases) {
testCases = in.nextInt();
}
while (testCases-- > 0) {
int n = in.nextInt();
int[] a = new int[n + 1];
for (int i = 1; i <= n; i++) a[i] = in.nextInt();
int count = 0;
// for (int i = 1; i <= n; i++) {
// for (int j = a[i] - i; j <= n; j += a[i]) {
// if (j >= 0) {
// if (i + j == a[i] * a[j] && i < j) count++;
// }
// }
// }
for (int i = 1; i <= n; i++) {
for (int j = a[i] - i; j <= n; j += a[i]) {
if (j >= 0) {
if ((long)a[i] * a[j] == (long)i + j && i < j) {
count++;
}
}
}
}
sb.append(count).append(newLine);
}
out.print(sb);
out.flush();
}
static class InputReader {
private final InputStream stream;
private final byte[] buf = new byte[8192];
private int curChar, snumChars;
private SpaceCharFilter filter;
public InputReader(InputStream stream) {
this.stream = stream;
}
public int snext() {
if (snumChars == -1) throw new InputMismatchException();
if (curChar >= snumChars) {
curChar = 0;
try {
snumChars = stream.read(buf);
} catch (IOException e) {
throw new InputMismatchException();
}
if (snumChars <= 0) return -1;
}
return buf[curChar++];
}
public int nextInt() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
int res = 0;
do {
if (c < '0' || c > '9') throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public long nextLong() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
int sgn = 1;
if (c == '-') {
sgn = -1;
c = snext();
}
long res = 0;
do {
if (c < '0' || c > '9') throw new InputMismatchException();
res *= 10;
res += c - '0';
c = snext();
} while (!isSpaceChar(c));
return res * sgn;
}
public int[] nextIntArray(int n) {
int a[] = new int[n];
for (int i = 0; i < n; i++) {
a[i] = nextInt();
}
return a;
}
public String readString() {
int c = snext();
while (isSpaceChar(c)) {
c = snext();
}
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isSpaceChar(c));
return res.toString();
}
public String nextLine() {
int c = snext();
while (isSpaceChar(c)) c = snext();
StringBuilder res = new StringBuilder();
do {
res.appendCodePoint(c);
c = snext();
} while (!isEndOfLine(c));
return res.toString();
}
public boolean isSpaceChar(int c) {
if (filter != null) return filter.isSpaceChar(c);
return c == ' ' || c == '\n' || c == '\r' || c == '\t' || c == -1;
}
private boolean isEndOfLine(int c) {
return c == '\n' || c == '\r' || c == -1;
}
public interface SpaceCharFilter {
boolean isSpaceChar(int ch);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 35b427b5c04cc5786f6024320ff2fd38 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | /* package codechef; // don't place package name! */
import java.util.*;
import java.lang.*;
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
public class Codechef
{
public static void main (String[] args) throws java.lang.Exception
{
// your code goes here
BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
int t = Integer.parseInt(br.readLine());
for(int q=0;q<t;q++){
int n = Integer.parseInt(br.readLine());
int a[] = new int[n];
String s[] = br.readLine().split(" ");
for(int i=0;i<n;i++){
a[i] = Integer.parseInt(s[i]);
}
int count=0;
for(int i=0;i<n;i++){
for(int j=1;j<2*n+1;j++){
int tmp = a[i]*j -i-1;
// System.out.println(tmp+" "+i);
if(tmp>n){
break;
}
else if(tmp<=0){
continue;
}
else{
if(a[tmp-1]==j && i>=tmp){
count++;
}
}
}
}
System.out.println(count);
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | c779768ca94e34f695b17b495dee2c96 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | /*
⠀⠀⠀⠀⣠⣶⡾⠏⠉⠙⠳⢦⡀⠀⠀⠀⢠⠞⠉⠙⠲⡀⠀
⠀⠀⠀⣴⠿⠏⠀⠀⠀⠀⠀⠀⢳⡀⠀⡏⠀⠀Y⠀⠀⢷
⠀⠀⢠⣟⣋⡀⢀⣀⣀⡀⠀⣀⡀⣧⠀⢸⠀⠀A⠀⠀ ⡇
⠀⠀⢸⣯⡭⠁⠸⣛⣟⠆⡴⣻⡲⣿⠀⣸⠀⠀S⠀ ⡇
⠀⠀⣟⣿⡭⠀⠀⠀⠀⠀⢱⠀⠀⣿⠀⢹⠀⠀H⠀⠀ ⡇
⠀⠀⠙⢿⣯⠄⠀⠀⠀⢀⡀⠀⠀⡿⠀⠀⡇⠀⠀⠀⠀⡼
⠀⠀⠀⠀⠹⣶⠆⠀⠀⠀⠀⠀⡴⠃⠀⠀⠘⠤⣄⣠⠞⠀
⠀⠀⠀⠀⠀⢸⣷⡦⢤⡤⢤⣞⣁⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀
⠀⠀⢀⣤⣴⣿⣏⠁⠀⠀⠸⣏⢯⣷⣖⣦⡀⠀⠀⠀⠀⠀⠀
⢀⣾⣽⣿⣿⣿⣿⠛⢲⣶⣾⢉⡷⣿⣿⠵⣿⠀⠀⠀⠀⠀⠀
⣼⣿⠍⠉⣿⡭⠉⠙⢺⣇⣼⡏⠀⠀⠀⣄⢸⠀⠀⠀⠀⠀⠀
⣿⣿⣧⣀⣿………⣀⣰⣏⣘⣆⣀⠀⠀
*/
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter; // System.out is a PrintStream
// import java.util.Arrays;
// import java.util.ArrayDeque;
// import java.util.ArrayList;
// import java.util.Collections; // for sorting ArrayList mainly
// import java.util.HashMap;
// import java.util.TreeMap;
// import java.util.HashSet;
import java.util.TreeSet;
// import java.util.Random;
import java.util.StringTokenizer;
public class B1541 {
public static void main(String[] args) throws IOException {
FastScanner scn = new FastScanner();
PrintWriter out = new PrintWriter(System.out);
for (int tc = scn.nextInt(); tc > 0; tc--) {
int N = scn.nextInt(), count = 0, sz = N * 2;
int[] arr = new int[N];
for (int i = 0; i < N; i++) {
arr[i] = scn.nextInt();
}
for (int i = 0; i < N; i++) {
long val = arr[i];
for (int j = 1; j * val <= sz; j++) {
int idx = (int)(val * j - (i + 1));
if (i + 1 < idx && idx <= N && arr[idx - 1] == j) {
count++;
}
}
}
out.println(count);
}
out.close();
}
/* ------------------- Sorting ------------------- */
private static void ruffleSort(int[] arr) {
// int N = arr.length;
// Random rand = new Random();
// for (int i = 0; i < N; i++) {
// int oi = rand.nextInt(N), temp = arr[i];
// arr[i] = arr[oi];
// arr[oi] = temp;
// }
// Arrays.sort(arr);
}
/* ------------------- Pair class ------------------- */
private static class Pair implements Comparable<Pair> {
int x, y;
Pair(int x, int y) {
this.x = x;
this.y = y;
}
public int compareTo(Pair o) {
return this.x == o.x ? this.y - o.y : this.x - o.x;
}
}
/* ------------------- HCF and LCM ------------------- */
private static int gcd(int num1, int num2) {
int temp = 0;
while (num2 != 0) {
temp = num1;
num1 = num2;
num2 = temp % num2;
}
return num1;
}
private static int lcm(int num1, int num2) {
return (int)((1L * num1 * num2) / gcd(num1, num2));
}
/* ------------------- primes and prime factorization ------------------- */
private static boolean[] seive(int N) {
// true means not prime, false means is a prime number :)
boolean[] notPrimes = new boolean[N + 1];
notPrimes[0] = notPrimes[1] = true;
for (int i = 2; i * i <= N; i++) {
if (notPrimes[i]) continue;
for (int j = i * i; j <= N; j += i) {
notPrimes[j] = true;
}
}
return notPrimes;
}
/* private static TreeMap<Integer, Integer> primeFactors(long N) {
TreeMap<Integer, Integer> primeFact = new TreeMap<>();
for (int i = 2; i <= Math.sqrt(N); i++) {
int count = 0;
while (N % i == 0) {
N /= i;
count++;
}
if (count != 0) {
primeFact.put(i, count);
}
}
if (N != 1) {
primeFact.put((int)N, 1);
}
return primeFact;
} */
/* ------------------- Binary Search ------------------- */
private static long factorial(int N) {
long ans = 1L;
for (int i = 2; i <= N; i++) {
ans *= i;
}
return ans;
}
private static long[] factorialDP(int N) {
long[] factDP = new long[N + 1];
factDP[0] = factDP[1] = 1;
for (int i = 2; i <= N; i++) {
factDP[i] = factDP[i - 1] * i;
}
return factDP;
}
private static long factorialMod(int N, int mod) {
long ans = 1L;
for (int i = 2; i <= N; i++) {
ans = ((ans % mod) * (i % mod)) % mod;
}
return ans;
}
/* ------------------- Binary Search ------------------- */
private static int lowerbound(int[] arr, int st, int ed, int tar) {
int ans = -1;
while (st <= ed) {
int mid = ((ed - st) >> 1) + st;
if (arr[mid] <= tar) {
ans = mid;
st = mid + 1;
} else {
ed = mid - 1;
}
}
return ans;
}
private static int upperbound(int[] arr, int st, int ed, int tar) {
int ans = -1;
while (st <= ed) {
int mid = ((ed - st) >> 1) + st;
if (arr[mid] < tar) {
st = mid + 1;
} else {
ans = mid;
ed = mid - 1;
}
}
return ans;
}
/* ------------------- Power Function ------------------- */
public static long pow(long x, long y) {
long res = 1;
while (y > 0) {
if ((y & 1) != 0) {
res = (res * x);
y--;
}
x = (x * x);
y >>= 1;
}
return res;
}
public static long powMod(long x, long y, int mod) {
long res = 1;
while (y > 0) {
if ((y & 1) != 0) {
res = (res * x) % mod;
y--;
}
x = (x * x) % mod;
y >>= 1;
}
return res % mod;
}
/* ------------------- Disjoint Set(Union and Find) ------------------- */
private static class DSU {
public int[] parent, size;
DSU(int n) {
parent = new int[n];
for (int i = 0; i < n; i++) {
parent[i] = i;
size[i] = 1;
}
}
public int find(int a) {
if (parent[a] == a) {
return a;
}
return parent[a] = find(parent[a]);
}
public void merge(int a, int b) {
int parA = find(a), parB = find(b);
if (parA == parB) return;
if (size[parA] < size[parB]) {
parent[parB] = parA;
size[parA] += size[parB];
} else {
parent[parA] = parB;
size[parB] += size[parA];
}
}
}
/* ------------------- Scanner class for input ------------------- */
private static class FastScanner {
BufferedReader br;
StringTokenizer st;
FastScanner() {
this.br = new BufferedReader(new InputStreamReader(System.in));
this.st = new StringTokenizer("");
}
public String next() {
while (!st.hasMoreTokens()) {
try {
st = new StringTokenizer(br.readLine());
} catch (IOException err) {
err.printStackTrace();
}
}
return st.nextToken();
}
public String nextLine() {
if (st.hasMoreTokens()) {
return st.nextToken("").trim();
}
try {
return br.readLine().trim();
} catch (IOException err) {
err.printStackTrace();
}
return "";
}
public int nextInt() {
return Integer.parseInt(next());
}
public long nextLong() {
return Long.parseLong(next());
}
public double nextDouble() {
return Double.parseDouble(next());
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 8dff26219b52df4b1c0552827e614d45 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.DataInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
public class pairs{
static class pair{
long ele;
long pos;
pair(long x,long y){
ele=x;
pos=y;
}
long getEle(){
return ele;
}
}
// static class ELEComparator implements Comparator{
// public int compare(Object o1,Object o2){
// pair s1=(pair) o1;
// pair s2=(pair)o2;
// if(s1.ele>s2.ele)
// return 1;
// else
// return -1;
// }
// }
public static void main(final String[] args) throws IOException {
long t=console.nextLong();
while(t-->0){
int n =console.nextInt();
long item;
long count=0;
pair[] B=new pair[n];
for(int i=0;i<n;i++){
item=console.nextLong();
B[i]=new pair(item,i+1);
}
Comparator<pair> cm1=Comparator.comparing(pair::getEle);
Arrays.sort(B,cm1);
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(B[i].ele *B[j].ele >= 2* n) break;
if(B[i].ele*B[j].ele==(B[i].pos+B[j].pos))
count++;
}
}
System.out.println(count);
}
}
private static final pairs.InputReader console = new pairs.InputReader();
static final class InputReader {
private final int BUFFER_SIZE = 1 << 24;
private final DataInputStream in = new DataInputStream(System.in);
private final byte[] buffer = new byte[BUFFER_SIZE];
private final byte[] buf = new byte[BUFFER_SIZE];
private int bufferPointer = 0, bytesRead = 0;
private void fillBuffer() throws IOException {
bytesRead = in.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) {
buffer[0] = -1;
}
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) {
fillBuffer();
}
return buffer[bufferPointer++];
}
public String next() throws IOException {
int cnt = 0, c;
while ((c = read()) != -1) {
if (c < 33) {
break;
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public String nextLine() throws IOException {
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\r' || c == '\n') {
break;
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
return (int) nextLong();
}
public long nextLong() throws IOException {
long ret = 0L, c = read();
while (c < 33L) {
c = read();
}
boolean neg = (c == 45L);
if (neg) {
c = read();
}
do {
ret = (ret << 3) + (ret << 1) + c - 48L;
} while ((c = read()) > 47 && c < 58);
return (neg ? -ret : ret);
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public void close() throws IOException {
in.close();
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | a846ee1317d68ad92717a6e7223c5bfa | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import java.io.DataInputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.util.*;
public class pairs{
static class pair{
long ele;
long pos;
pair(long x,long y){
ele=x;
pos=y;
}
long getEle(){
return ele;
}
}
// static class ELEComparator implements Comparator{
// public int compare(Object o1,Object o2){
// pair s1=(pair) o1;
// pair s2=(pair)o2;
// if(s1.ele>s2.ele)
// return 1;
// else
// return -1;
// }
// }
public static void main(final String[] args) throws IOException {
long t=console.nextLong();
while(t-->0){
int n =console.nextInt();
long item;
long count=0;
pair[] B=new pair[n];
for(int i=0;i<n;i++){
item=console.nextLong();
B[i]=new pair(item,i+1);
}
Comparator<pair> cm1=Comparator.comparing((pair x)->x.ele);
Arrays.sort(B,cm1);
for(int i=0;i<n;i++){
for(int j=i+1;j<n;j++){
if(B[i].ele *B[j].ele >= 2L * n) break;
if(B[i].ele*B[j].ele==(B[i].pos+B[j].pos))
count++;
}
}
System.out.println(count);
}
}
private static final pairs.InputReader console = new pairs.InputReader();
static final class InputReader {
private final int BUFFER_SIZE = 1 << 24;
private final DataInputStream in = new DataInputStream(System.in);
private final byte[] buffer = new byte[BUFFER_SIZE];
private final byte[] buf = new byte[BUFFER_SIZE];
private int bufferPointer = 0, bytesRead = 0;
private void fillBuffer() throws IOException {
bytesRead = in.read(buffer, bufferPointer = 0, BUFFER_SIZE);
if (bytesRead == -1) {
buffer[0] = -1;
}
}
private byte read() throws IOException {
if (bufferPointer == bytesRead) {
fillBuffer();
}
return buffer[bufferPointer++];
}
public String next() throws IOException {
int cnt = 0, c;
while ((c = read()) != -1) {
if (c < 33) {
break;
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public String nextLine() throws IOException {
int cnt = 0, c;
while ((c = read()) != -1) {
if (c == '\r' || c == '\n') {
break;
}
buf[cnt++] = (byte) c;
}
return new String(buf, 0, cnt);
}
public int nextInt() throws IOException {
return (int) nextLong();
}
public long nextLong() throws IOException {
long ret = 0L, c = read();
while (c < 33L) {
c = read();
}
boolean neg = (c == 45L);
if (neg) {
c = read();
}
do {
ret = (ret << 3) + (ret << 1) + c - 48L;
} while ((c = read()) > 47 && c < 58);
return (neg ? -ret : ret);
}
public double nextDouble() throws IOException {
return Double.parseDouble(next());
}
public void close() throws IOException {
in.close();
}
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 1a97e9d7a7e7ed68e5a092149c709099 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;
import java.util.Map.Entry;
public class MacroTemplates
{
public static void main(String hi[]) throws Exception
{
PrintWriter out = new PrintWriter(System.out);
FastReader ob=new FastReader();
int T = ob.nextInt();
StringBuilder sb = new StringBuilder();
while(T-->0)
{
int n = ob.nextInt();
int arr[]=new int[n];
for(int i=0;i<n;i++)
{
arr[i]=ob.nextInt();
}
int[] map = new int[2*n+1];
for(int i=0; i < n; i++)
map[arr[i]] = i+1;
int res = 0;
for(int v=1; v <= 2*n; v++)
if(map[v] > 0)
{
//ai = v
//aj = x/v
for(int x=v; x <= 2*n; x+=v)
if(map[x/v] > 0 && map[v]+map[x/v] == x && v != x/v)
res++;
}
out.println(res/2);
out.flush();
}
}
public static boolean isPrime(long n)
{
if(n < 2) return false;
if(n == 2 || n == 3) return true;
if(n%2 == 0 || n%3 == 0) return false;
long sqrtN = (long)Math.sqrt(n)+1;
for(long i = 6L; i <= sqrtN; i += 6) {
if(n%(i-1) == 0 || n%(i+1) == 0) return false;
}
return true;
}
public static long gcd(long a, long b)
{
if(a > b)
a = (a+b)-(b=a);
if(a == 0L)
return b;
return gcd(b%a, a);
}
public static ArrayList<Integer> findDiv(int N)
{
//gens all divisors of N
ArrayList<Integer> ls1 = new ArrayList<Integer>();
ArrayList<Integer> ls2 = new ArrayList<Integer>();
for(int i=1; i <= (int)(Math.sqrt(N)+0.00000001); i++)
if(N%i == 0)
{
ls1.add(i);
ls2.add(N/i);
}
Collections.reverse(ls2);
for(int b: ls2)
if(b != ls1.get(ls1.size()-1))
ls1.add(b);
return ls1;
}
public static void sort(int[] arr)
{
//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static long power(long x, long y, long p)
{
//0^0 = 1
long res = 1L;
x = x%p;
while(y > 0)
{
if((y&1)==1)
res = (res*x)%p;
y >>= 1;
x = (x*x)%p;
}
return res;
}
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output | |
PASSED | 24b9a2246656f4d742fa268d596356f9 | train_107.jsonl | 1624635300 | You are given an array $$$a_1, a_2, \dots, a_n$$$ consisting of $$$n$$$ distinct integers. Count the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | 256 megabytes | import static java.lang.Math.max;
import static java.lang.Math.min;
import static java.lang.Math.abs;
import static java.lang.System.out;
import java.util.*;
import java.io.*;
import java.math.*;
import java.util.Map.Entry;
public class MacroTemplates
{
public static void main(String hi[]) throws Exception
{
PrintWriter out = new PrintWriter(System.out);
FastReader ob=new FastReader();
int T = ob.nextInt();
StringBuilder sb = new StringBuilder();
while(T-->0)
{
int n = ob.nextInt();
int a[]=new int[n];
for(int i=0;i<n;i++)
{
a[i]=ob.nextInt();
}
int c=0;
Map<Integer,Integer> arr=new TreeMap<>();
for(int i=0;i<n;i++)
{
arr.put(a[i], i+1);
}
for(int i=0;i<n;i++)
{
Set<Entry<Integer,Integer>> entries=arr.entrySet();
for(Entry<Integer,Integer> entry : entries)
{
if(a[i]!=entry.getKey())
{
if((i+1+entry.getValue()) == a[i]*entry.getKey())
{
c++;
}
if(a[i]*entry.getKey() >= 2*n)
{
break;
}
}
}
}
out.println(c/2);
out.flush();
}
}
public static boolean isPrime(long n)
{
if(n < 2) return false;
if(n == 2 || n == 3) return true;
if(n%2 == 0 || n%3 == 0) return false;
long sqrtN = (long)Math.sqrt(n)+1;
for(long i = 6L; i <= sqrtN; i += 6) {
if(n%(i-1) == 0 || n%(i+1) == 0) return false;
}
return true;
}
public static long gcd(long a, long b)
{
if(a > b)
a = (a+b)-(b=a);
if(a == 0L)
return b;
return gcd(b%a, a);
}
public static ArrayList<Integer> findDiv(int N)
{
//gens all divisors of N
ArrayList<Integer> ls1 = new ArrayList<Integer>();
ArrayList<Integer> ls2 = new ArrayList<Integer>();
for(int i=1; i <= (int)(Math.sqrt(N)+0.00000001); i++)
if(N%i == 0)
{
ls1.add(i);
ls2.add(N/i);
}
Collections.reverse(ls2);
for(int b: ls2)
if(b != ls1.get(ls1.size()-1))
ls1.add(b);
return ls1;
}
public static void sort(int[] arr)
{
//because Arrays.sort() uses quicksort which is dumb
//Collections.sort() uses merge sort
ArrayList<Integer> ls = new ArrayList<Integer>();
for(int x: arr)
ls.add(x);
Collections.sort(ls);
for(int i=0; i < arr.length; i++)
arr[i] = ls.get(i);
}
public static long power(long x, long y, long p)
{
//0^0 = 1
long res = 1L;
x = x%p;
while(y > 0)
{
if((y&1)==1)
res = (res*x)%p;
y >>= 1;
x = (x*x)%p;
}
return res;
}
}
class FastReader {
BufferedReader br;
StringTokenizer st;
public FastReader()
{
br = new BufferedReader(
new InputStreamReader(System.in));
}
String next()
{
while (st == null || !st.hasMoreElements()) {
try {
st = new StringTokenizer(br.readLine());
}
catch (IOException e) {
e.printStackTrace();
}
}
return st.nextToken();
}
int nextInt() { return Integer.parseInt(next()); }
long nextLong() { return Long.parseLong(next()); }
double nextDouble()
{
return Double.parseDouble(next());
}
String nextLine()
{
String str = "";
try {
str = br.readLine();
}
catch (IOException e) {
e.printStackTrace();
}
return str;
}
}
| Java | ["3\n2\n3 1\n3\n6 1 5\n5\n3 1 5 9 2"] | 2 seconds | ["1\n1\n3"] | NoteFor the first test case, the only pair that satisfies the constraints is $$$(1, 2)$$$, as $$$a_1 \cdot a_2 = 1 + 2 = 3$$$For the second test case, the only pair that satisfies the constraints is $$$(2, 3)$$$.For the third test case, the pairs that satisfy the constraints are $$$(1, 2)$$$, $$$(1, 5)$$$, and $$$(2, 3)$$$. | Java 11 | standard input | [
"brute force",
"implementation",
"math",
"number theory"
] | 0ce05499cd28f0825580ff48dae9e7a9 | The first line contains one integer $$$t$$$ ($$$1 \leq t \leq 10^4$$$) — the number of test cases. Then $$$t$$$ cases follow. The first line of each test case contains one integer $$$n$$$ ($$$2 \leq n \leq 10^5$$$) — the length of array $$$a$$$. The second line of each test case contains $$$n$$$ space separated integers $$$a_1, a_2, \ldots, a_n$$$ ($$$1 \leq a_i \leq 2 \cdot n$$$) — the array $$$a$$$. It is guaranteed that all elements are distinct. It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$2 \cdot 10^5$$$. | 1,200 | For each test case, output the number of pairs of indices $$$(i, j)$$$ such that $$$i < j$$$ and $$$a_i \cdot a_j = i + j$$$. | standard output |
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