[bzoj5339][TJOI2018]教科书般的亵渎【拉格朗日插值法】

【题目链接】
　　https://www.lydsy.com/JudgeOnline/problem.php?id=5339
　　https://loj.ac/problem/2578
【题解】
　　显然$k=m+1$，然后就是一个$k$次幂的前缀和减去几个没有的值。
　　用插值法可以实现。
　　时间复杂度$O(N^3)$
【代码】

# 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;
}
ll readll(){
    ll 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;
ll num[N], n;
int m, x[N], y[N];
int power(int x, int y){
    int i = x; x = 1;
    while (y > 0){
        if (y % 2 == 1) x = 1ll * x * i % P;
        i = 1ll * i * i % P;
        y /= 2;
    }
    return x;
}
int getsum(ll tmp, int lim){
    int sum = 0; tmp %= P;
    for (int i = 1; i <= lim; i++){
        int num1 = 1, num2 = 1;
        for (int j = 1; j <= lim; j++)
            if (i != j){
                num1 = 1ll * num1 * (tmp - 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);
    for (int opt = read(); opt--;){
        n = readll(), m = read();
        for (int i = 1; i <= m; i++)
            num[i] = readll();
        sort(num + 1, num + m + 1);
        int k = m + 1;
        for (int i = 1; i <= k + 2; i++)
            x[i] = i, y[i] = (y[i - 1] + power(i, k)) % P; 
        int ans = 0;
        for (int i = 1; i <= k; i++){
            ll now = n - num[i - 1];
            ans = (ans + getsum(now, k + 2)) % P;
            for (int j = i; j <= m; j++)
                ans = (ans - power(num[j] - num[i - 1], k)) % P;
        }
        printf("%d\n", (ans + P) % P);
    }
    return 0;
}