prompt string | response string |
|---|---|
Problem: 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 ... | 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 ... |
Problem: 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 ... | #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] =... |
Problem: 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 ... | #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... |
Problem: 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 ... | 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);
/... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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;... |
Problem: 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 ... | #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 =... |
Problem: 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 ... | #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 / ... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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");... |
Problem: 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 ... | 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)]... |
Problem: 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 ... | 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)... |
Problem: 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 ... | #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)... |
Problem: 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 ... | #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) ... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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 ... |
Problem: 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 ... | #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... |
Problem: 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 ... | 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... |
Problem: 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 ... | #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 (... |
Problem: 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 ... | #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;
... |
Problem: 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 ... | #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(... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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) {
... |
Problem: 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 ... | #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 -... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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) {
... |
Problem: 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 ... | #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)... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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 =... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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;... |
Problem: 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 ... | #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 (... |
Problem: 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 ... | 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 = []
... |
Problem: 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 ... | #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') {
... |
Problem: 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 ... | #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) {
... |
Problem: 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 ... | #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) {
... |
Problem: 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 ... | #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... |
Problem: 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 ... | 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... |
Problem: 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 ... | 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();
... |
Problem: 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 ... | 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... |
Problem: 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 ... | #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()... |
Problem: 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 ... | 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)
... |
Problem: 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 ... | 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;... |
Problem: 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 ... | 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... |
Problem: 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 ... | 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... |
Problem: 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 ... | 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... |
Problem: 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 ... | #------------------------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... |
Problem: 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 ... | 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... |
Problem: 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 ... | #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()) {
... |
Problem: 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 ... | #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 ... |
Problem: 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 ... | #!/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 ''... |
Problem: 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 ... | #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++;
... |
Problem: 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 ... | 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... |
Problem: 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 ... | #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 >= ... |
Problem: 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 ... | #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++;
... |
Problem: 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 ... | #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... |
Problem: 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 ... | 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... |
Problem: 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 ... | 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]... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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'... |
Problem: 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 ... | #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' ... |
Problem: 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 ... | #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() ... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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 ... |
Problem: 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 ... | /**
* 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();
... |
Problem: 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 ... |
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... |
Problem: 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 ... | #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;
... |
Problem: 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 ... | #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 ... |
Problem: 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 ... | #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... |
Problem: 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 ... | 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 = ... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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");
... |
Problem: 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 ... |
/*
* 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) {
... |
Problem: 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 ... | 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;... |
Problem: 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 ... | #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++) {
... |
Problem: 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 ... | #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 (... |
Problem: 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 ... | #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 ... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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;
... |
Problem: 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 ... | #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;
... |
Problem: 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 ... | """
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):
... |
Problem: 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 ... | #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++) {... |
Problem: 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 ... | 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... |
Problem: 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 ... | 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... |
Problem: 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 ... | 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 ... |
Problem: 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 ... | 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;
... |
Problem: 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 ... | 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... |
Problem: 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 ... | #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++) {
... |
Problem: 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 ... | 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(" ")).... |
Problem: 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 ... | #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... |
Problem: 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 ... | 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... |
Problem: 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 ... | #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... |
Problem: 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 ... | #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])) {
... |
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