Search is not available for this dataset
name stringlengths 2 112 | description stringlengths 29 13k | source int64 1 7 | difficulty int64 0 25 | solution stringlengths 7 983k | language stringclasses 4
values |
|---|---|---|---|---|---|
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | def ng():print("NO");exit()
n, k = map(int,input().split())
if (n,k)==(1,0):print("YES");print(0);exit()
ans = [0] * n;u = 0;popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4): ans[v] = v // 2+1
k ... | PYTHON3 |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 7;
int s[MAXN];
int isComplete(int n) { return (n & (n + 1)) == 0; }
int gen(int n, int niche) {
for (int i = 2; i <= niche; i += 2) {
s[i] = i / 2;
s[i + 1] = i / 2;
}
int last = 1;
for (int i = niche + 1; i <= n; i += 2) {
s[i] =... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int ans[100010];
void solve(const int &l, const int &r, const int &k);
inline char legal(const int &N, const int &k) {
if (N < 0 || k < 0) return false;
char flag;
int n = N + 1 >> 1;
if (n == 1)
flag = !k;
else if (n == 5)
flag = (k == 1 || k == 3);
els... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | import java.io.*;
import java.util.*;
public class Solution {
int bound = 100000;
int N = 31;
private void solve() throws Exception {
memo = new Boolean[N+1][N+1];
nodes = new Node[N+1][N+1];
// for(int i = 1; i <= n; i+=2) {
// String a = String.format("%3d", i);
/... | JAVA |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
inline int read() {
char ch = getchar();
int nega = 1;
while (!isdigit(ch)) {
if (ch == '-') nega = -1;
ch = getchar();
}
int ans = 0;
while (isdigit(ch)) {
ans = ans * 10 + ch - 48;
ch = getchar();
}
if (nega == -1) return -ans;
return ans... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 100009;
int judge(int n, int k) {
if (k < 0) return 0;
if (n % 2 == 0) return 0;
if (k == 0) {
if (__builtin_popcount(n + 1) == 1)
return 1;
else
return 0;
}
if (k == 1) {
if (__builtin_popcount(n + 1) == 1)
return 0;... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e15 + 7;
bool check(long long n) { return __builtin_popcount(n + 1) == 1; }
vector<long long> res;
bool solve(long long n, long long k, long long par) {
if (n == 11 && k =... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n, k;
inline int lowbit(int x) { return x & -x; }
void get_lian(int num) {
printf("0 ");
for (int i = 1; i <= num; i++) {
printf("%d %d ", i * 2 - 1, i * 2 - 1);
}
}
void get_binary(int root, int num) {
for (int i = 2; i <= num; i++) {
printf("%d ", i / ... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
int cnt;
int par[MAXN], br[MAXN];
void solve(int n, int k, int root) {
if (n == 0) return;
int curr = ++cnt;
par[curr] = root;
if (k == 0) {
solve((n - 1) / 2, 0, curr);
solve((n - 1) / 2, 0, curr);
} else if (k == 1) {
int pot... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n, k, fa[100005];
bool check(int x) { return x - (x & -x) == 0; }
int main() {
scanf("%d%d", &n, &k);
int mx = max((n - 3) / 2, 0);
if (k > mx || (n % 2 == 0) || (n == 9 && k == 2) ||
(check(n + 1) && k == 1) || (!check(n + 1) && k == 0)) {
printf("NO");... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | from heapq import *
import sys
int1 = lambda x: int(x) - 1
p2D = lambda x: print(*x, sep="\n")
def II(): return int(sys.stdin.readline())
def MI(): return map(int, sys.stdin.readline().split())
def LI(): return list(map(int, sys.stdin.readline().split()))
def LLI(rows_number): return [LI() for _ in range(rows_number)]... | PYTHON3 |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | def ng():
print("NO")
exit()
n, k = map(int,input().split())
if (n,k)==(1,0):
print("YES")
print(0)
exit()
ans = [0] * n
popcnt = lambda x: bin(x).count("1")
if n & 1 == 0 or n < 2 * k + 3 or (popcnt(n + 1) > 1 and k == 0): ng()
u = 0
if popcnt(n + 1 - 2 * (k - 1)) == 1:
for v in range(1, 4)... | PYTHON3 |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int mx(int u) {
if (u < 5) return 0;
return (u - 3) / 2;
}
bool can(int u, int v) {
if (u == 9 && v == 2) {
return false;
}
if ((u & 1) == 0) {
return false;
}
if (v == 0) {
return __builtin_popcount(u + 1) == 1;
}
if (v > 1) {
return mx(u)... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int fa[100005];
void js() {
puts("NO");
exit(0);
}
int check(int x) { return x == (x & -x); }
int main() {
int n, k;
scanf("%d%d", &n, &k);
if (!(n & 1)) js();
if (k > max(0, (n - 3) / 2)) js();
if (check(n + 1) && k == 1) js();
if (!check(n + 1) && k == 0) ... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
const int maxn = 1e5 + 5;
const int pow2 = 1 << 20;
int fa[maxn], N;
int cnt = 1;
void dfs(int o, int n, int k) {
if (k <= 1) {
for (int i = o + 1; i <= n; i++) fa[i] = (i - o + 1) / 2 + o - 1;
if (k == 1 && pow2 % (n - o + 2) == 0) {
fa[n] = N;
fa[n - 1] = N;
}
re... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
template <class TH>
void _dbg(const char *sdbg, TH h) {
cerr << sdbg << '=' << h << endl;
}
template <class TH, class... TA>
void _dbg(const char *sdbg, TH h, TA... a) {
while (*sdbg != ',') cerr << *sdbg++;
cerr << '=' << h << ',';
_dbg(sdbg + 1, a...);
}
template ... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
inline bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline int msbp(int x) { return 31 - __builtin_clz(x); }
inline int msb(int x) { return 1 << msbp(x); }
const int INF = 0x3f3f3f3f;
const int N = 1e5 + 10;
void bad() {
cout << "NO" << endl;
exit(0);
}
in... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
import java.io.BufferedReader;
import java.io.InputStream;
/**
* Built using CHelper plug-in
* Actual soluti... | JAVA |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
const int bufSize = 1e6;
inline char nc() {
static char buf[bufSize], *p1 = buf, *p2 = buf;
return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, bufSize, stdin), p1 == p2)
? EOF
: *p1++;
}
template <typename T>
inline T read(T &r) {
static char c;
r = 0;
for (... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n, k;
inline int lowbit(int x) { return x & (-x); }
bool check() {
if (n % 2 == 0) return 0;
if (max((n - 3) / 2, 0) < k) return 0;
if (n == 9 && k == 2) return 0;
if (n + 1 == lowbit(n + 1) && k == 1) return 0;
if (n + 1 != lowbit(n + 1) && k == 0) return 0;
... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 7;
int ans[maxn];
int lowbit(int x) { return x & -x; }
bool is_two(int x) { return x == lowbit(x); }
bool dfs(int n, int k, int id) {
if (k == 1) {
printf("%d %d %d\n", n, k, id);
if (is_two(n)) {
return false;
}
ans[id] = max(... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int MAXN = 2e5 + 7;
bool isbalanced(int x, int y) { return (2 * x > y && 2 * y > x); }
bool ispow(int x) { return (__builtin_popcount(x + 1) == 1); }
int getdep(int x) { return __builtin_ctz(x + 1); }
int n, k, cnt;
int ans[MAXN];
void build(int fa, int dep) {
if (d... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int cur;
bool possible(int n, int k) {
if (n % 2 == 0) return false;
if ((n & (n + 1)) == 0) {
if (k == 0)
return true;
else if (k == 1)
return false;
}
if (n == 9 && k == 2) return false;
return k > 0 && k < n / 2;
}
void dfs(int n, int k, int... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int memo[105][6];
pair<int, int> next1[105][6];
pair<int, int> next2[105][6];
bool boleh(int n, int k);
int dp(int n, int k) {
if (n % 2 == 0) {
return false;
}
if (k < 0) {
return false;
}
if (n == 1) {
return k == 0;
}
if (memo[n][k] != -1) {
... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int pow2 = 1 << 20;
int fa[maxn], N;
int cnt = 1;
void dfs(int o, int n, int k) {
if (k <= 1) {
for (int i = o + 1; i <= n; i++) fa[i] = (i - o + 1) / 2 + o - 1;
if (k == 1 && pow2 % (n - o + 2) == 0) {
fa[n] = N;
fa[n -... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
namespace ztd {
using namespace std;
template <typename T>
inline T read(T& t) {
t = 0;
short f = 1;
char ch = getchar();
double d = 0.1;
while (ch < '0' || ch > '9') {
if (ch == '-') f = -f;
ch = getchar();
}
while (ch >= '0' && ch <= '9') t = t * 10 + ch - '0', ch = getc... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
bool can(long long n, long long k) {
if (n % 2 == 0) return 0;
if (k == 0) return !(n & (n + 1));
if (k == 1) return n & (n + 1);
return (k > 0 && k * 2 + 3 <= n) && !(k == 2 && n == 9);
}
long long i = 1;
void construct(long long n, long long k, long long a = 0) {
... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int n, k, w;
void v(int a) { cout << a << " "; }
void t(int a, int b) {
for (int i = a + 1; i <= b; i++) v(a + (i - a - 1) / 2);
}
void s(int db, int p, int f) {
v(f);
if (db < 2)
t(p, n);
else {
if (db == 3 && n - p == 10)
t(p, p + 6), t(p + 6, p + 8)... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
template <class T>
inline void cmin(T &a, T b) {
((a > b) && (a = b));
}
template <class T>
inline void cmax(T &a, T b) {
((a < b) && (a = b));
}
char IO;
template <class T = int>
T rd() {
T s = 0;
int f = 0;
while (!isdigit(IO = getchar())) f |= IO == '-';
do s... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int ans[100011];
int n, k;
inline int ok(int n, int k) {
if (n % 2 == 0) return 0;
if (n == 1 and k == 0) return 1;
if (k > n / 2 - 1) return 0;
if (n == 9 and k == 2) return 0;
if (__builtin_popcount(n + 1) == 1) {
if (k == 1) return 0;
return 1;
}
re... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
int ans[100011];
int n, k;
inline int ok(int n, int k) {
if (n % 2 == 0) return 0;
if (n == 1 and k == 0) return 1;
if (k > n / 2 - 1) return 0;
if (n == 9 and k == 2) return 0;
if (__builtin_popcount(n + 1) == 1) {
if (k == 1) return 0;
return 1;
}
re... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
bool ok(int n, int k) {
if (n % 2 == 0) return false;
if (n == 1) return k == 0;
bool pw2 = ((n + 1) & n) == 0;
if (pw2 && k == 1) return false;
if (!pw2 && k == 0) return false;
if (k > (n - 3) / 2) return false;
if (n == 9 && k == 2) return false;
return t... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
const long long inf = 1e15 + 7;
bool check(long long n) { return __builtin_popcount(n + 1) == 1; }
vector<long long> res;
bool solve(long long n, long long k, long long par) {
if (n == 11 && k =... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int inf = (int)1e9;
const long long INF = (long long)5e18;
const int MOD = 998244353;
int _abs(int x) { return x < 0 ? -x : x; }
int add(int x, int y) {
x += y;
return x >= MOD ? x - MOD : x;
}
int sub(int x, int y) {
x -= y;
return x < 0 ? x + MOD : x;
}
void... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
#pragma GCC optimize("O3")
#pragma GCC target("sse4")
using namespace std;
template <class T>
bool ckmin(T& a, const T& b) {
return b < a ? a = b, 1 : 0;
}
template <class T>
bool ckmax(T& a, const T& b) {
return a < b ? a = b, 1 : 0;
}
mt19937 rng(chrono::steady_clock::now().time_since_epo... | CPP |
1379_E. Inverse Genealogy | Ivan is fond of genealogy. Currently he is studying a particular genealogical structure, which consists of some people. In this structure every person has either both parents specified, or none. Additionally, each person has exactly one child, except for one special person, who does not have any children. The people in... | 2 | 11 | #include <bits/stdc++.h>
using namespace std;
const int nax = 100;
int mem[nax][nax];
int ok(int n, int k) {
if (n % 2 == 0 || n <= 0 || k < 0) return 0;
if (n == 1) return (k == 0);
int& memo = mem[n][k];
if (!memo) {
int r = 0;
for (int na = 1; na < n - 1; na = na * 2 + 1) {
int ka = 0;
in... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
typedef struct {
char c[15000];
int ipos, len;
} st;
bool glas(char s) {
if (s == 'a' || s == 'e' || s == 'i' || s == 'o' || s == 'u') return true;
return false;
}
int findPos(st* a, int k) {
if (a->len == -1) a->len = strlen(a->c);
if (a->ipos == -2) return -1;... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const double PI = 4 * atan(1.0);
void fast_stream() { std::ios_base::sync_with_stdio(0); }
int K;
bool isVowel(char a) {
return a == 'a' || a == 'i' || a == 'u' || a == 'e' || a == 'o';
}
bool match(string &a, string &b) {
int cnt = 0;
for (int i = 0;; i++) {
if (... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | n, k = [int(it) for it in raw_input().split()]
Quatrains = []
Results = []
for i in range(0, n):
temp = []
temp.append(raw_input())
temp.append(raw_input())
temp.append(raw_input())
temp.append(raw_input())
Quatrains.append(temp)
def CheckQuatrain(a):
def Suffix(a):
result = []
... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int style[2500];
int n, k;
string lk[4];
string getLaskK(const string& str) {
if (str.size() < k) return "";
int l = str.size() - 1;
int tot = 0;
while (l >= 0) {
if (str[l] == 'a' || str[l] == 'e' || str[l] == 'i' || str[l] == 'o' ||
str[l] == 'u') {
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
string s[4];
int t, n, k;
int pd[4][4];
map<char, int> mp;
char ss;
bool xd(string s1, string s2) {
int l1, l2;
int tt = 0;
l1 = s1.size() - 1;
l2 = s2.size() - 1;
while (l1 >= 0 && tt < k) {
if (mp[s1[l1]]) {
tt++;
}
l1--;
}
if (tt < k) {
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, k;
int a[2510];
int ans;
int main() {
cin >> n >> k;
for (int i = 0; i < n; i++) {
string s[4];
string p[4];
for (int j = 0; j < 4; j++) {
cin >> s[j];
int t = s[j].size() - 1;
int cur = 0;
while (t >= 0 && cur != k) {
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long long Maxn = 1e5 + 7;
const long long Max = 1e3 + 7;
const long long Mod = 1e9 + 7;
const long long Inf = 1e9 + 7;
string s[Maxn], V[Maxn];
long long ans[5];
long long check(char t) {
if (t == 'u' || t == 'o' || t == 'a' || t == 'i' || t == 'e') return true;
r... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.*;
import java.lang.*;
import java.io.*;
import java.awt.Point;
// U KNOW THAT IF THIS DAY WILL BE URS THEN NO ONE CAN DEFEAT U HERE................
// ASCII = 48 + i ;// 2^28 = 268,435,456 > 2* 10^8 // log 10 base 2 = 3.3219
// odd:: (x^2+1)/2 , (x^2-1)/2 ; x>=3// even:: (x^2/4)+1 ,(x^2/4)-1 x >=4... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.*;
import java.lang.*;
import java.math.*;
import java.io.*;
import static java.lang.Math.*;
import static java.util.Arrays.*;
public class C{
Scanner sc=new Scanner(System.in);
int INF=1<<28;
double EPS=1e-9;
int n, k;
String[] ss;
void run(){
n=sc.nextInt();
... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | from sys import stdin
rints = lambda: [int(x) for x in stdin.readline().split()]
str_inp = lambda n: [stdin.readline()[:-1] for x in range(n)]
n, k = rints()
a, vow, pat, flag = [str_inp(4) for _ in range(n)], 'aeiou', set(), 0
pats = {'aabb', 'abab', 'abba'}
for i in range(n):
ixs = [0] * (4)
for j in range... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int countbit(int n) { return (n == 0) ? 0 : (1 + countbit(n & (n - 1))); }
int lowbit(int n) { return (n ^ (n - 1)) & n; }
const double pi = acos(-1.0);
const double eps = 1e-11;
char str[4][10010];
int N, K;
char mode[4][5] = {"aaaa", "aabb", "abba", "abab"};
int getMode()... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import sys
n,k=map(int,sys.stdin.readline().split())
def func(s):
if (s==''):
return
cur_gl=k
pos=len(s)-1
while pos>=0 and cur_gl>0:
if s[pos] in ('a','e','i','o','u'):
cur_gl-=1
if cur_gl>0:
pos-=1
if cur_gl>0:
print 'NO'
exit(0)
... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.Scanner;
// C
public class Main {
private static boolean isVowel(char c) {
return c=='a' || c=='e' || c=='i' || c=='o' || c=='u';
}
public static void main (String[] args) {
Scanner c = new Scanner(System.in);
int n = c.nextInt();
int k = c.nextInt();
c.nextLine();
boolean aabb = true;... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | 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 soluti... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import static java.lang.Math.*;
import java.util.*;
import java.io.*;
public class C {
public void solve() throws Exception {
int n = nextInt(), kk = nextInt();
boolean aabb=true, abab=true, abba=true;
for (int i=0; i<n; ++i) {
String[] q = new String[4];
int[] cnt... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.io.*;
import java.util.*;
public class Main implements Runnable{
private void solve()throws IOException{
HashSet<Character> vowels=new HashSet<>();
vowels.add('a');
vowels.add('e');
vowels.add('i');
vowels.add('o');
vowels.add('u');
int n=nextInt();
int... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #------------------------template--------------------------#
import os
import sys
from math import *
from collections import *
from fractions import *
from bisect import *
from heapq import*
from io import BytesIO, IOBase
def vsInput():
sys.stdin = open('input.txt', 'r')
sys.stdout = open('output.txt', 'w')
BUF... | PYTHON3 |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.*;
public class c {
public static void main(String[] args)
{
Scanner input = new Scanner(System.in);
int n = input.nextInt();
int k = input.nextInt();
String a, b, c, d;
boolean abab = true, abba = true, aabb = true, aaaa = true;
for(int i = 0; i < n; i++)
{
a = inpu... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const string tp[4] = {"aaaa", "aabb", "abab", "abba"};
int n, k;
string w[10000];
int main() {
cin >> n >> k;
n *= 4;
bool bad = false;
for (int i = 0; i < n; i++) {
string s;
cin >> s;
w[i] = "";
int l = 0;
while (l < k && (int)s.size()) {
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int k;
bool Is(char a) {
return (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u');
}
bool Equal(string s, string t) {
int n = (int)s.size(), m = (int)t.size();
int v1 = 0, v2 = 0;
for (int i = 0; i < n; i++) v1 += Is(s[i]);
for (int i = 0; i < m; i++) v2 ... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #!/usr/bin/env python
"""(c) gorlum0 [at] gmail.com"""
import itertools as it
from sys import stdin
vowels = 'aeiou'
types_rhymes = 'NO abba abab # aabb # # aaaa'.split()
def suffix(k, w):
for i in xrange(len(w)-1, -1, -1):
if w[i] in vowels: k -= 1
if not k: break
return w[i:] if not k else ''... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int k;
bool eq(char &q) {
return (q == 'a' || q == 'e' || q == 'o' || q == 'u' || q == 'i');
}
bool rifm(string &a, string &b) {
int n = a.length(), m = b.length();
int k1, k2;
k1 = k2 = 0;
int i = n;
while (i > 0 && k1 < k) {
i--;
if (eq(a[i])) k1++;
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | def bad(a,b,c,d):
if a == 'bad' or b == 'bad' or c == 'bad' or d == 'bad':
return True
return False
if __name__ == "__main__":
p = map(eval,raw_input().split(" "))
n=p[0]
K=p[1]
poem = []
vowel = ['a','e','i','o','u']
for i in range(0,n):
quat = []
for j in range... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, n, k, l, m;
bool a, b, c;
string s, t[4];
a = b = c = true;
cin >> n >> k;
for ((i) = 0; (i) < (int)(n); (i)++) {
for ((j) = 0; (j) < (int)(4); (j)++) {
cin >> s;
l = 0;
t[j] = "";
for (m = s.length() - 1; m >= ... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, k;
int rhymes[2550];
string arr[4];
int vowel(char a) {
return (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u');
}
int rhyme(string a, string b) {
int cnt = 0;
int i;
for ((i) = 0; (i) < (int)(((int)(a).size())); (i)++)
if (vowel(a[i])) cnt++;
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int i, j, k;
int n, m;
while (cin >> n >> m) {
int f = 0;
for (i = 0; i < n; ++i) {
string t[4];
for (j = 0; j < 4; ++j) {
string s;
cin >> s;
if (f < 0) continue;
int ln = s.length();
int ct = 0... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.StringTokenizer;
public class Main {
static String[] list;
static String type(String s[]){
if(s[0].equals(s[1]) && s[1].equals(s[2]) && s[2].equals(s[3])) return "aaaa";
if(s[0].equal... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.*;
import java.lang.*;
public class Main
{
public static boolean rhymes(String[][] ap, int a, int b, int k) {
int la = ap[a].length, lb = ap[b].length;
//System.out.println(la);
for(int i = la - k, j = lb - k; i < la && j < lb;) {
if(i < 1 || j < 1) return false;
//System.out.println(ap[a]... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<int> aux;
int n, k;
char m[100][150];
bool vogal(char a) {
return (a == 'a' || a == 'e' || a == 'i' || a == 'o' || a == 'u');
}
bool igual(string a, int ida, string b, int idb) {
if (a.length() - ida != b.length() - idb) return false;
for (int i = ida; i < a.le... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
string S1[10002], an[] = {"aabb", "abab", "abba", "aaaa", "NO"}, S;
int ar[5], n, m;
int main() {
cin >> n >> m;
for (int i = 0; i < 4 * n; i++) {
int l = 0, k;
cin >> S;
for (k = S.size(); k >= 0 && l != m;) {
k--;
if (S[k] == 'a' || S[k] == 'e'... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const long double EPS = 1E-9;
const int INF = (int)1E9;
const long long INF64 = (long long)1E18;
int n, k;
bool check1(string a, string b, string c, string d) {
string p, q, r, s;
int t = 0;
for (int i = a.size() - 1; i >= 0; i--) {
if (a[i] == 'a' || a[i] == 'e' ... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
set<char> V = {'a', 'e', 'i', 'o', 'u'};
bool match(string& a, string& b, int k) {
int i = a.size() - 1, j = b.size() - 1;
while (i >= 0 && j >= 0) {
if (a[i] != b[j]) return 0;
if (V.count(a[i]) && --k == 0) return 1;
i--, j--;
}
return 0;
}
int main() ... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxi = 1e6 + 10;
int a[maxi];
string s[maxi];
int n, k;
vector<int> v[maxi];
int vowel(char x) {
return x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u';
}
int rima(int x, int y) {
int n1 = s[x].size();
int n2 = s[y].size();
int pozX, pozY;
int v... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
string lines[10001];
string pieces[10001];
int n, k;
char v[] = {'a', 'e', 'i', 'o', 'u'};
int main() {
cin >> n >> k;
for (int i = 0; i < n; i++) {
cin >> lines[4 * i] >> lines[4 * i + 1] >> lines[4 * i + 2] >>
lines[4 * i + 3];
}
set<char> vowels(v, v ... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | /**
* Jerry Ma
* Program lit
*/
import java.io.*;
import java.util.*;
public class lit {
static BufferedReader cin;
static StringTokenizer tk;
static char [] vowels = {'a','e','i','o','u'};
public static void main (String [] args) throws IOException {
init();
int num = gInt(), vowelPos = gInt();
... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 |
import java.util.HashMap;
import java.util.Scanner;
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
/**
*
* @author Alex
*/
public class _139C {
private static String getSuffix(String a, int k) {
String res = "";
StringBuilder temp = new Stri... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int N = 100005;
string s[N], suf[N];
string ans[4] = {"aaaa", "aabb", "abab", "abba"};
int main() {
int n, k;
scanf("%d%d", &n, &k);
for (int i = 1; i <= n; ++i) {
for (int j = (i - 1) * 4 + 1; j <= i * 4; ++j) {
cin >> s[j];
}
}
int f = -1;
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool pro(char c) {
return (c == 'a' || c == 'i' || c == 'o' || c == 'e' || c == 'u');
}
long long n, k;
bool ttt(string a, string b) {
long long t = min(a.length(), b.length()), i = 1, u = 0;
while (i <= t) {
if (a[a.length() - i] == b[b.length() - i]) {
if ... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int inf = 1 << 30, N = 1002;
int ax[] = {0, 1, -1, 0, 0};
int ay[] = {0, 0, 0, -1, 1};
string a[10002];
bool rifma(int first, int second, int k) {
int n = a[first].size() - 1;
int m = a[second].size() - 1;
while (true) {
if (a[first][n] != a[second][m]) retu... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
int n = in.nextInt();
int k = in.nextInt();
String cur = "";
String tmp;
int cnt = 0;
String s[] = new String[4];
for (int i = ... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ii = pair<int, int>;
using vi = vector<int>;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3fLL;
const double err = 1e-9;
const int mod = 1e9 + 7;
const int N = 1e5 + 10;
ll n, k;
bool isVowel(char c) {
return c == 'a' or c == 'e' o... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k, i, j, l;
string s, t, r, z, u, p[5], q;
char c;
cin >> n >> k;
getline(cin, z);
for (i = 1; i <= n; i++) {
t = " ";
r = "";
c = 'a';
for (l = 1; l <= 4; l++) {
getline(cin, z);
j = z.length() - 1;
q = s... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
string getBack(string s, int k) {
int p = s.size();
for (int i = 0; i < k; i++) {
int ap = s.substr(0, p).find_last_of("a");
int ep = s.substr(0, p).find_last_of("e");
int ip = s.substr(0, p).find_last_of("i");
int op = s.substr(0, p).find_last_of("o");
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 |
/*
* To change this template, choose Tools | Templates
* and open the template in the editor.
*/
import java.util.HashSet;
import java.util.Scanner;
/**
*
* @author Eslam Ashraf
*/
public class C {
static HashSet<Integer> set = new HashSet<Integer>();
public static void main(String[] args) {
... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.*;
import java.math.*;
public class main
{
public static void main(String args[])
{
Scanner c=new Scanner(System.in);
int N=c.nextInt();
int K=c.nextInt();
String P[][]=new String [N][4];
for(int i=0;i<N;i++)
{
for(int j=0;... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
vector<int> room;
string str[4];
int check(char x) {
if (x == 'a' || x == 'e' || x == 'i' || x == 'u' || x == 'o') return 1;
return 0;
}
int main() {
int n, m;
scanf("%d%d", &n, &m);
int ret = 0;
for (int i = 0; i < n; i++) {
for (int j = 0; j < 4; j++) {
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
void nope() {
cout << "NO";
exit(0);
}
int main() {
set<string> res;
int n, k;
cin >> n >> k;
for (int i = 0; i < n; i++) {
string sp[4];
for (int j = 0; j < 4; j++) {
string s;
cin >> s;
int p = s.size();
int c = 0;
while (... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
const int maxn = 10010;
string s1[maxn];
string check(string t1, string t2, string t3, string t4) {
if (t1 == t2 && t3 == t4 && t1 == t3) return "aaaa";
if (t1 == t3 && t2 == t4 && t1 != t2) return "abab";
if (t1 == t4 && t2 == t3 && t1 != t2) return "abba";
if (t1 ... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int mode(char a[], char b[], char c[], char d[]) {
if (strcmp(a, b) == 0 && strcmp(c, d) == 0 && strcmp(b, c) != 0)
return 1;
else if (strcmp(a, b) == 0 && strcmp(c, d) == 0 && strcmp(a, c) == 0)
return 0;
else if (strcmp(a, c) == 0 && strcmp(b, d) == 0 && str... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
char str[4][10005];
int k;
bool isV(char ch) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u') return 1;
return 0;
}
int get_type() {
int len[4], i, j, st[4], x;
for (i = 0; i < 4; i++) {
len[i] = strlen(str[i]);
if (len[i] < k) return 0;
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int main() {
int n, k;
cin >> n >> k;
char c[5] = {'a', 'e', 'i', 'o', 'u'};
int r[4] = {0};
int flag = 0;
for (int i = 0; i < n; i++) {
vector<string> v;
for (int i = 0; i < 4; i++) {
string str, p = "";
cin >> str;
int ctr = 0;
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | """
Literature Lesson
"""
import random
def split_str(string, k):
index = len(string)
for char in string[::-1]:
index -= 1
if char in 'aeiou':
k -= 1
if k == 0:
return string[index:]
return string * random.randint(0, 100)
def rhyme_schemes(quatrain, k):
... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
int n, k;
char a[4][100000];
int c[4];
int b[3000];
bool is(char ch) {
if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
return true;
return false;
}
int i, j, m;
int main() {
scanf("%d%d", &n, &k);
bool flag = true;
for (i = 1; i <= n; i++) {... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import sys
def finish(result):
print(result)
sys.exit()
n, k = map(int, raw_input().split())
scheme_count = { 'aabb': 0, 'abab': 0, 'abba': 0, 'aaaa': 0 }
def make_scheme(suffixes):
if None in suffixes:
return None
suffix_hash = {}
symbol = 'a'
scheme_parts = []
for suffix in suffixes:
if suffi... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.InputStreamReader;
import java.io.FileNotFoundException;
import java.io.FileReader;
import java.io.FileWriter;
import java.io.IOException;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.util.StringTokenizer;
publ... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import re
r = lambda: raw_input().strip()
n,k = map(int,r().split(' '))
poem = [r().strip()[::-1] for i in xrange(4*n)]
counter = 0
ptype = 'aaaa'
s = []
for line in poem:
i = k
suffix = None
match = re.search('([^aeiou]*[aeiou]){%d}' % k, line)
if match:
suffix = match.group(0)
if suffix ... | PYTHON |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.*;
import java.math.*;
public class codeforces {
public static int go(Scanner sc,int k)
{
String[] data=new String[4];
for(int i=0;i<4;i++)
{
data[i]=sc.next();
int t=0;
... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.io.*;
import java.util.*;
public class CF139C {
static boolean vowel(char c) {
return c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u';
}
static String suffix(String s, int k) {
for (int i = s.length() - 1; i >= 0; i--)
if (vowel(s.charAt(i))) {
k--;
if (k == 0)
return s.subs... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
int main(void) {
int type = 0, i, j, k, l, m, n;
char str[4][10001], last[4][10001];
scanf("%d %d%*c", &n, &k);
while (n--) {
for (i = 0; i < 4; i++) scanf("%s%*c", str[i]);
for (m = 0; m < 4; m++) {
l = strlen(str[m]);
for (i = 0, j = 0; i < l && j < k; i++) {
... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.util.Scanner;
public class LiteratureLesson {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
String s = in.nextLine();
int n = Integer.parseInt(s.substring(0, s.indexOf(" ")).trim());
int k = Integer.parseInt(s.substring(s.indexOf(" ")).... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
struct sd {
int f, s, t;
};
string ff(string a, int k) {
string tt = "";
for (int i = a.size() - 1; i >= 0; i--) {
if (k == 0) break;
if (a[i] == 'a' || a[i] == 'e' || a[i] == 'i' || a[i] == 'o' || a[i] == 'u')
k--;
tt += a[i];
}
if (k != 0) retu... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Scanner;
public class codefo {
public static void main(String[] args) throws IOException
{
Scanner scan = new Scanner(System.in);
scan.useDelimiter("\\p{javaWhitespace}+|[\\s,]+");
/*int n = Integer.v... | JAVA |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool is[10020][5];
int n, k;
bool Is(char x) {
if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u')
return 1;
else
return 0;
}
char *solve(char *x) {
int t = strlen(x), c = 0;
for (int i = t; i >= 0; --i) {
if (Is(x[i])) {
c++;
if (c... | CPP |
139_C. Literature Lesson | Vera adores poems. All the poems Vera knows are divided into quatrains (groups of four lines) and in each quatrain some lines contain rhymes.
Let's consider that all lines in the poems consist of lowercase Latin letters (without spaces). Letters "a", "e", "i", "o", "u" are considered vowels.
Two lines rhyme if their ... | 2 | 9 | #include <bits/stdc++.h>
using namespace std;
bool f[2510][5];
int n, k;
bool Is(char x) {
if (x == 'a' || x == 'e' || x == 'i' || x == 'o' || x == 'u') return 1;
return 0;
}
bool NO;
char *judge(char *c) {
if (NO) return NULL;
int l = strlen(c), t = 0;
for (int i = l - 1; i >= 0; --i) {
if (Is(c[i])) {
... | CPP |
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