# | Time | Username | Problem | Language | Result | Execution time | Memory |
---|---|---|---|---|---|---|---|
22833 | AcornCkiGs14004Team (#40) | hi (KRIII5_HI) | C++14 | 3 ms | 2104 KiB |
This submission is migrated from previous version of oj.uz, which used different machine for grading. This submission may have different result if resubmitted.
#include <stdio.h>
#include <algorithm>
#include <assert.h>
#include <cmath>
#include <complex>
#include <deque>
#include <functional>
#include <iostream>
#include <limits.h>
#include <map>
#include <math.h>
#include <queue>
#include <set>
#include <stdlib.h>
#include <string.h>
#include <string>
#include <time.h>
#include <unordered_map>
#include <unordered_set>
#include <vector>
#pragma warning(disable:4996)
#pragma comment(linker, "/STACK:336777216")
using namespace std;
#define mp make_pair
#define Fi first
#define Se second
#define pb(x) push_back(x)
#define szz(x) ((int)(x).size())
#define rep(i, n) for(int i=0;i<n;i++)
#define all(x) (x).begin(), (x).end()
#define ldb ldouble
typedef tuple<int, int, int> t3;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
typedef pair <ll, int> pli;
typedef pair <db, db> pdd;
int IT_MAX = 1 << 19;
const ll MOD = 1000000007;
const int INF = 0x3f3f3f3f;
const ll LL_INF = 1034567890123456789ll;
const db PI = acos(-1);
const db ERR = 1E-11;
int in[105];
int dp[105][105][2];
int main() {
int N, i, j, k;
scanf("%d", &N);
if (N != 2) return 0;
scanf("%d %d", &in[1], &in[2]);
for (i = 1; i <= 100; i++) dp[i][0][1] = i;
for (i = 1; i <= 100; i++) {
for (j = 1; j <= 100; j++) {
for (k = 1; k <= i; k++) dp[i][j][1] = (dp[i][j][1] + dp[i - k][j][0] * k) % MOD;
for (k = 1; k <= j; k++) dp[i][j][0] = (dp[i][j][0] + dp[i][j - k][1] * k) % MOD;
}
}
int ans = 0;
for (i = 0; i <= in[1] - 1; i++) {
for (j = 0; i+j <= in[1] - 1; j++) {
ans = (ans + dp[in[2]][in[1] - i - j - 1][1] * (i+j+1)) % MOD;
}
}
for (i = 0; i <= in[2] - 1; i++) {
for (j = 0; i + j <= in[2] - 1; j++) {
ans = (ans + dp[in[1]][in[2] - i - j - 1][1] * (i + j + 1)) % MOD;
}
}
return !printf("%d\n", ans);
}
Compilation message (stderr)
# | Verdict | Execution time | Memory | Grader output |
---|---|---|---|---|
Fetching results... |
# | Verdict | Execution time | Memory | Grader output |
---|---|---|---|---|
Fetching results... |