| # |
Submission time |
Handle |
Problem |
Language |
Result |
Execution time |
Memory |
| 218879 |
2020-04-03T00:51:39 Z |
eggag32 |
Zapina (COCI20_zapina) |
C++17 |
|
5 ms |
384 KB |
#pragma GCC optimize ("O3")
#pragma GCC target ("sse4")
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef pair<int, int> pi;
#define debug(x) cerr << #x << ": " << x << endl;
#define debug2(x, y) debug(x) debug(y);
#define repn(i, a, b) for(int i = (int)(a); i < (int)(b); i++)
#define rep(i, a) for(int i = 0; i < (int)(a); i++)
#define all(v) v.begin(), v.end()
#define mp make_pair
#define pb push_back
#define lb lower_bound
#define ub upper_bound
#define fi first
#define se second
#define sq(x) ((x) * (x))
const int MOD = 1e9 + 7;
template<class T> T gcd(T a, T b){ return ((b == 0) ? a : gcd(b, a % b)); }
ll dp[355][355][2]; //[index][tot][already have one happy]
ll fact[355];
ll invFact[355];
ll C(ll n, ll r){
if(r > n) return 0;
ll ret = fact[n];
ret = ((ret * invFact[n - r]) + MOD) % MOD;
ret = ((ret * invFact[r]) + MOD) % MOD;
return ret;
}
ll mypow(ll a, ll p){
if(p == 0) return 1;
if(p == 1) return (a % MOD);
if(p & 1) return ((a % MOD) * (mypow(a, p - 1) % MOD)) % MOD;
ll x = mypow(a, p / 2) % MOD;
return (x * x) % MOD;
}
ll inv(ll base){
return mypow(base, MOD - 2);
}
int main(){
ios_base::sync_with_stdio(false);
cin.tie(0);
//freopen("input.in", "r", stdin);
//freopen("output.out", "w", stdout);
int n;
cin >> n;
rep(i, n + 2) rep(j, n + 2) rep(k, 2) dp[i][j][k] = 0;
fact[0] = 1;
invFact[0] = 1;
for(ll i = 1; i < 355; i++){
fact[i] = (1LL * fact[i - 1] * i) % MOD;
invFact[i] = inv(fact[i]);
}
dp[1][1][1] = n;
repn(i, 0, n + 1) if(i != 1) dp[1][i][0] = C(n, i);
repn(i, 2, n + 1){
repn(j, 0, n + 1){ //the prev total
repn(k, 0, n + 1) if((j + k) <= n){ //current choice
if(k == i){
dp[i][j + k][1] = (((dp[i][j + k][1] + dp[i - 1][j][1] + dp[i - 1][j][0]) % MOD) * C(n - j, k)) % MOD;
dp[i][j + k][0] = (((dp[i][j + k][0] + dp[i - 1][j][0]) % MOD) * C(n - j, k)) % MOD;
}
else{
dp[i][j + k][1] = (((dp[i][j + k][1] + dp[i - 1][j][1]) % MOD) * C(n - j, k)) % MOD;
dp[i][j + k][0] = (((dp[i][j + k][0] + dp[i - 1][j][0]) % MOD) * C(n - j, k)) % MOD;
}
}
}
}
cout << dp[n][n][1] % MOD << endl;
return 0;
}
/*
Things to look out for:
- Integer overflows
- Array bounds
- Special cases
Be careful!
*/
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
5 ms |
384 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
5 ms |
384 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
| # |
Verdict |
Execution time |
Memory |
Grader output |
| 1 |
Incorrect |
5 ms |
384 KB |
Output isn't correct |
| 2 |
Halted |
0 ms |
0 KB |
- |
