Codeforces 233 D

D - Table

思路：dp

首先，第i列的个数肯定和第i - n列个数一样，假设[i - n + 1, i - 1] 之间的个数之和为x，那么第i列和第i-n列的个数应该是n - x

那么我们可以用dp求方案数

状态：dp[i][j] 表是到第 i 列为止 填了 j 个的方案数

初始状态： dp[0][0] = 1

状态转移： dp[i][j](1 <= i <= n, 0 <= j <= k) = ∑(dp[i-1][j - l](l <= n && j >= l) * C(n, l) ^ cnt) 

其中，cnt = n/m 或者 n/m + 1，C(n, l)^cnt 可以预处理来降低复杂度

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize(4)
#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define pi acos(-1.0)
#define LL long long
//#define mp make_pair
#define pb push_back
#define ls rt<<1, l, m
#define rs rt<<1|1, m+1, r
#define ULL unsigned LL
#define pll pair<LL, LL>
#define pii pair<int, int>
#define piii pair<pii, int>
#define mem(a, b) memset(a, b, sizeof(a))
#define fio ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define fopen freopen("in.txt", "r", stdin);freopen("out.txt", "w", stout);
//head

const int MOD = 1e9 + 7;
const int N = 110;
LL dp[N][N*N];
LL qp[N][N][2];
LL q_pow(LL n, LL k) {
    LL ans = 1;
    while(k) {
        if(k&1) ans = (ans * n) % MOD;
        n = (n * n) % MOD;
        k >>= 1;
    }
    return ans;
}
int main() {
    int n, k;
    LL m;
    scanf("%d %lld %d", &n, &m, &k);
    LL cnt = m/n;
    LL C = 1;
    for (int i = 0; i <= n; i++) {
        qp[n][i][0] = q_pow(C, cnt);
        qp[n][i][1] = q_pow(C, cnt+1);
        C = (C * (n-i)) % MOD;
        C = (C * q_pow(i+1, MOD-2)) % MOD;
    }
    dp[0][0] = 1;
    for (int i = 1; i <= n; i++) {
        for (int j = 0; j <= k; j++) {
            for (int l = 0; l <= n && j >= l; l++) {
                if(i <= m%n) dp[i][j] = (dp[i][j] + dp[i-1][j-l] * qp[n][l][1]) % MOD;
                else dp[i][j] = (dp[i][j] + dp[i-1][j-l] * qp[n][l][0]) % MOD;
            }
        }
    }
    printf("%lld
", dp[n][k]);
    return 0;
}