51nod1228 序列求和(自然数幂和)

与UVA766 Sum of powers类似，见http://www.cnblogs.com/IMGavin/p/5948824.html

由于结果对MOD取模，使用逆元

#include<cstdio>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<string>
#include<algorithm>
#include<map>
#include<queue>
#include<vector>
#include<cmath>
#include<utility>
using namespace std;
typedef long long LL;
const int N = 2016, INF = 0x3F3F3F3F, MOD = 1000000007;

LL bo[N];
LL cm[N][N], inv[N];

void init(){
    inv[1] = 1;
    for(int i = 2; i < N; i++){
    	inv[i] = (MOD - MOD / i ) * inv[MOD % i] % MOD;
    }

    memset(cm, 0, sizeof(cm));
    cm[0][0] = 1;
    for(int i = 1; i < N; i++){
        cm[i][0] = 1;
        for(int j = 1; j <= i; j++){
            cm[i][j] = (cm[i - 1][j - 1] + cm[i - 1][j]) % MOD;
        }
    }

    bo[0] = 1;
    for(int i = 1; i < N; i++){
        bo[i] = 0;
        for(int j = 0; j < i; j++){
        	bo[i] += cm[i + 1][j] * bo[j] % MOD;
        	bo[i] %= MOD;
        }
        bo[i] = (-bo[i] * inv[i + 1] % MOD + MOD) % MOD;
    }
    bo[1] = inv[2];
}

LL PowMod(LL a,LL b,LL MOD){//快速幂
    LL ret=1;
    while(b){
        if(b&1) ret=(ret*a)%MOD;
        a=(a*a)%MOD;
        b>>=1;
    }
    return ret;
}

LL solve(LL n, LL m){
	LL ans = 0;
	for(LL k = 0; k <= m; k++){
		ans += (cm[m + 1][k] * bo[k] % MOD) * PowMod(n % MOD, m + 1 - k, MOD) % MOD;
		ans %= MOD;
	}
	ans = ans * inv[m + 1] % MOD;
	return ans;
}

int main(){
    init();
    int t;
    cin >> t;
    while(t--){
        LL n, k;
        scanf("%I64d %I64d", &n, &k);
        printf("%I64d\n", solve(n, k));
    }
    return 0;
}