[bzoj4559][loj2026][JLoi2016]成績比較【拉格朗日插值法】【dp】

【題目鏈接】
　　https://www.lydsy.com/JudgeOnline/problem.php?id=4559
　　https://loj.ac/problem/2026
【題解】
　　記fi,j$f_{i,j}$ 表示當前統計到第i$i$ 門課程，仍然有j$j$ 個人被碾壓，可以枚舉通過上一門課程的碾壓人數來轉移。
　　記gi$g_i$ 表示第i$i$ 門課可行的方案數。
　　fi,j=∑nt=jfi−1,t∗C(n−1−t,ranki−1−(t−j))∗C(t,j)∗gi$f_{i,j}=\sum _{t=j}^n{f}_{i-1,t}\ast C(n-1-t,ran{k}_i-1-(t-j))\ast C(t,j)\ast g_i$
　　第一個組合數計算的是其他不被碾壓的人組合的方案數，第二個是選擇那些人被碾壓的方案數。
　　gi=∑Uii=1in−ranki∗(U−i)ranki−1$g_i=\sum _{i=1}^{U_i}{i}^{n-ran{k}_i}\ast (U-i{)}^{ran{k}_i-1}$
　　顯然這是一個n+1$n+1$ 次多項式，可以用插值法解決。
　　最後fm,k$f_{m,k}$ 即爲答案。
　　時間複雜度O(N3)$O(N^3)$
【代碼】

/* - - - - - - - - - - - - - - -
    User :      VanishD
    problem :
    Points :    
- - - - - - - - - - - - - - - */
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    N       110
using namespace std;
int read(){
    int tmp = 0, fh = 1; char ch = getchar();
    while (ch < '0' || ch > '9'){ if (ch == '-') fh = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9'){ tmp = tmp * 10 + ch - '0'; ch = getchar(); }
    return tmp * fh;
}
const int P = 1e9 + 7;
int n, m, k, lim[N], f[N][N], rk[N], c[N][N], num[N], x[N], y[N]; 
int power(int x, int y){
    int i = x; x = 1;
    while (y > 0){
        if (y % 2 == 1) x = 1ll * i * x % P;
        i = 1ll * i * i % P;
        y /= 2;
    }
    return x;
}
int solve(int lim, int n, int m){
    int num = n + m + 2, sum = 0;
    for (int i = 1; i <= num; i++) 
        x[i] = i, y[i] = (y[i - 1] + 1ll * power(i, n) * power(lim - i, m)) % P;
    for (int i = 1; i <= num; i++){
        int num1 = 1, num2 = 1;
        for (int j = 1; j <= num; j++){
                if (i != j){
                    num1 = 1ll * num1 * (lim - x[j]) % P;
                    num2 = 1ll * num2 * (x[i] - x[j]) % P;
                }
        }
        sum = (sum + 1ll * num1 * y[i] % P * power(num2, P - 2)) % P;
    }
    return sum;
}
int main(){
//  freopen(".in", "r", stdin);
//  freopen(".out", "w", stdout);
    n = read() - 1, m = read(), k = read();
    for (int i = 1; i <= m; i++) lim[i] = read();
    for (int i = 1; i <= m; i++) rk[i] = read() - 1;
    c[0][0] = 1;
    for (int i = 1; i <= n; i++){
        c[i][0] = 1;
        for (int j = 1; j <= i; j++)
            c[i][j] = (c[i - 1][j - 1] + c[i - 1][j]) % P;
    }
    for (int i = 1; i <= m; i++) num[i] = solve(lim[i], n - rk[i], rk[i]);
    f[0][n] = 1;
    for (int i = 1; i <= m; i++)
        for (int j = k; j <= n; j++)
            for (int t = j; t <= n; t++){
                int num1 = t - j, num2 = j;
                if (num1 > rk[i] || num2 > n - rk[i]) continue;
                f[i][j] = (f[i][j] + 1ll * c[t][j] % P * c[n - t][rk[i] - num1] % P * num[i] % P * f[i - 1][t]) % P;
            }
    printf("%d\n", (f[m][k] + P) % P);
    return 0;
}