洛谷P4593 [TJOI2018]教科书般的*(拉格朗日插值)

题意

题目链接

sol

打出暴力不难发现时间复杂度的瓶颈在于求\(\sum_{i = 1}^n i^k\)

老祖宗告诉我们，这东西是个\(k\)次多项式，插一插就行了

// 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
*/