LightOJ - 1132 Summing up Powers 矩阵快速幂

题目大意：求(1^K + 2^K + 3K + … + N^K) % 2^32

解题思路：
借用别人的图

可以先打表，求出Cnm,用杨辉三角可以快速得到

#include<cstdio>
typedef unsigned long long ll;
const int N = 55;
const ll mod = (1LL << 32);
struct Matrix{
    ll mat[N][N];
}A, B, tmp;

ll n, num[N];
ll C[N][N];
int K;

void init2() {
    C[0][0] = 1;
    for(int i = 1; i <= 50; i++) {
        C[i][i] = C[i][0] = 1;
        for(int j = 1; j < i; j++)
            C[i][j] = (C[i-1][j-1] + C[i-1][j]) % mod;
    }
}

Matrix matMul(const Matrix &x, const Matrix &y) {
    for(int i = 0; i < K + 2; i++)
        for(int j = 0; j < K + 2; j++) {
            tmp.mat[i][j] = 0;
            for(int k = 0; k < K + 2; k++) {
                tmp.mat[i][j] = (tmp.mat[i][j] + x.mat[i][k] * y.mat[k][j]) % mod;
            }
        }
    return tmp;
}
void solve() {
    while(n) {
        if(n & 1)
            B = matMul(B,A);
        A = matMul(A,A);
        n >>= 1;
    }
}

void init() {
    for(int i = 0; i < K + 2; i++) 
        for(int j = 0; j < K + 2; j++) {
            B.mat[i][j] = A.mat[i][j] = 0;
            if(i == j)
                B.mat[i][j] = 1;
        }

    A.mat[0][0] = 1;
    for(int i = 1; i < K + 2; i++) 
        A.mat[i][0] = A.mat[i][1] = C[K][i-1];
    for(int i = 2; i < K + 2; i++)
        for(int j = i; j < K + 2; j++) {
            A.mat[j][i] = C[K-i+1][j-i];
        }
}

int main() {
    int test, cas = 1;
    scanf("%d", &test);
    init2();
    while(test--) {
        scanf("%lld%d", &n, &K);
        printf("Case %d: ", cas++);
        ll ans = ( (n % mod) * ( (n + 1) % mod) / 2) % mod;
        if(K == 1) {
            printf("%lld\n", ans);
            continue;
        }
        init();

        n--;
        solve();
        ans = 0;
        for(int i = 0; i < K + 2; i++)
            ans = (ans + B.mat[i][0] ) % mod;
        printf("%lld\n", ans);
    }
    return 0;
}