BZOJ3157 国王奇遇记——神奇的推式子

先膜一发Miskcoo，大佬的博客上多项式相关的非常全
原题戳我

题目大意

求
\[\sum\limits_{i=1}^{n}i^mm^i\]

题解

设一个函数\(f(i)=\sum\limits_{j=1}^{n}j^im^j\)
然后貌似用一个叫扰动法（感觉就是错位相消法）的东西，算一下
\[(m-1)f(i)=\sum\limits_{j=1}^{n+1}(j-1)^im^j-\sum\limits_{i=1}^{n}j^im^j=n^im^{n+1}-\sum\limits_{j=1}^{n}m^j[(j-1)^i-j^i]\]
其中，\((j-1)^i-j^i\)可以用一波二项式展开化为\(\sum\limits_{k=0}^{i-1}\binom{i}{k}(-1)^{i-k}j^k\)，回带可得
\[(m-1)f(i)=n^im^{n+1}-\sum\limits_{j=1}^{n}m^j\sum\limits_{k=0}^{i-1}\binom{i}{k}(-1)^{i-k}j^k\]\[=n^im^{n+1}-\sum\limits_{k=0}^{i-1}\binom{i}{k}(-1)^{i-k}\sum\limits_{j=1}^{n}j^km^j\]\[=n^im^{n+1}-\sum\limits_{k=0}^{i-1}\binom{i}{k}(-1)^{i-k}f(k)\]
然后就有了一个\(O(m^2)\)的递推做法，还有一个\(O(m)\)的，但看起来挺麻烦的，咕了
以下是代码，注意初值\(f(0)\)的设置还有\(m=1\)时的特判

#include <algorithm>
#include  <iostream>
#include   <cstdlib>
#include   <cstring>
#include    <cstdio>
#include    <string>
#include    <vector>
#include     <cmath>
#include     <ctime>
#include     <queue>
#include       <map>
#include       <set>

using namespace std;

#define ull unsigned long long
#define pii pair<int, int>
#define uint unsigned int
#define mii map<int, int>
#define lbd lower_bound
#define ubd upper_bound
#define INF 0x3f3f3f3f
#define IINF 0x3f3f3f3f3f3f3f3fLL
#define vi vector<int>
#define ll long long
#define mp make_pair
#define pb push_back
#define re register
#define il inline

#define MOD 1000000007
#define M 1000

int n, m;
int C[M+5][M+5];
int f[M+5];

int fpow(int x, int p) {
    int ret = 1;
    while(p) {
        if(p&1) ret = 1LL*ret*x%MOD;
        x = 1LL*x*x%MOD;
        p >>= 1;
    }
    return ret;
}

void init() {
    for(int i = 0; i <= m; ++i) C[i][0] = 1;
    for(int i = 1; i <= m; ++i)
        for(int j = 1; j <= i; ++j)
            C[i][j] = (C[i-1][j-1]+C[i-1][j])%MOD;
        f[0] = (1LL*(fpow(m, n+1)-1)*fpow(m-1, MOD-2)%MOD-1+MOD)%MOD;
}

int main() {
    scanf("%d%d", &n, &m);
    if(m == 1) {
        printf("%lld\n", 1LL*n*(n+1)%MOD*fpow(2, MOD-2)%MOD);
        return 0;
    }
    init();
    for(int i = 1; i <= m; ++i) { // 递推
        f[i] = 1LL*fpow(n, i)*fpow(m, n+1)%MOD;
        for(int j = 0; j <= i-1; ++j) {
            if((i-j)&1) f[i] = (f[i]-1LL*C[i][j]*f[j]%MOD)%MOD;
            else f[i] = (f[i]+1LL*C[i][j]*f[j]%MOD)%MOD;
        }
        f[i] = (1LL*f[i]*fpow(m-1, MOD-2)%MOD+MOD)%MOD;
    }
    printf("%d\n", f[m]);
        return 0;
}

转载于:https://www.cnblogs.com/dummyummy/p/10913073.html