题意

题目链接

Sol

想不到想不到。。

首先在不考虑每个人的真是成绩的情况下，设\(f[i][j]\)表示考虑了前\(i\)个人，有\(j\)个人被碾压的方案数

转移方程：$$f[i][j] = \sum_{k = j}^n f[i -1][k] C_{k}^{k - j} C_{N - k}^{r[i] - 1 - (k - j)} * g(i)$$

大概解释一下，枚举的\(k\)表示之前碾压了多少，首先我们凑出\(j\)个人继续碾压，也就是说会有\(k - j\)个人该课的分数比\(B\)爷高，那么这\(k\)个人我们已经考虑完了

接下来需要从剩下的\(N-k\)个人中，选出\(r[i] - 1 - (k - j)\)个人，保证他们的分数比B爷高

后面的\(g(i)\)表示的是吧\(1 \sim U_i\)的分数，分给\(N\)个人后，有\(R_i\)个人比B爷高的方案数

这个计算的时候可以直接枚举B爷的分数

\(g(k) = \sum_{i = 1}^{U_k} i^{N - R[i]} * (U_k - i) ^{R[i] - 1}\)

后面的次数小于等于\(N-1\)，然后直接插值一下就行了

// luogu-judger-enable-o2

#include<bits/stdc++.h>

using namespace std;

const int MAXN = 103, mod = 1e9 + 7;

inline int read() {

    char c = getchar(); int x = 0, f = 1;

    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}

    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();

    return x * f;

}

int add(int x, int y) {

    if(x + y < 0) return x + y + mod;

    return x + y >= mod ? x + y - mod : x + y;

}

int mul(int x, int y) {

    return 1ll * x * y % mod;

}

int fp(int a, int p) {

    int base = 1;

    while(p) {

        if(p & 1) base = mul(base, a);

        a = mul(a, a); p >>= 1;

    }

    return base;

}

int N, M, K, f[MAXN][MAXN], C[MAXN][MAXN], U[MAXN], R[MAXN], g[MAXN];

int get(int U, int R) {

    memset(g, 0, sizeof(g));

    for(int i = 1; i <= MAXN - 1; i++)

        for(int k = 1; k <= i; k++)

            g[i] = add(g[i], mul(fp(k, N - R), fp(i - k, R - 1)));

    int ans = 0;

    for(int i = 1; i <= MAXN - 1; i++) {

        int up = 1, down = 1;

        for(int j = 1; j <= MAXN - 1; j++) {

            if(i == j) continue;

            up = mul(up, add(U, -j));

            down = mul(down, add(i, -j));

        }

        ans = add(ans, mul(g[i], mul(up, fp(down, mod - 2))));

    }

    return ans;

}

int main() {

    //freopen("a.in", "r", stdin);

    N = read(); M = read(); K = read();

    for(int i = 0; i <= N; i++) {

    	C[i][0] = C[i][i] = 1;

    	for(int j = 1; j < i; j++) C[i][j] = add(C[i - 1][j - 1], C[i - 1][j]);

    }

    for(int i = 1; i <= M; i++) U[i] = read();

    for(int i = 1; i <= M; i++) R[i] = read();

    f[0][N - 1] = 1;

    for(int i = 1; i <= M; i++) {

    	int t = get(U[i], R[i]);

    	for(int j = K; j <= N; j++) {

    		for(int k = j; k <= N - 1; k++)

    			if(k - j <= R[i] - 1) f[i][j] = add(f[i][j], mul(mul(f[i - 1][k], C[k][k - j]), C[N - 1 - k][R[i] - 1 - (k - j)]));

    		f[i][j] = mul(f[i][j], t);

    	}

    }

    printf("%d", f[M][K]);

    return 0;

}

/*

100 3 50

500 500 456

13 46 45

*/