LightOJ

题意

Given N and K, you have to find

(1^K + 2^K + 3^K + … + N^K) % 2^32

思路

构造矩阵，快速幂。
进行二项展开后得到杨辉三角。

链接

https://vjudge.net/contest/176567#problem/D

代码

#include<cstdio>#include<iostream>#include<vector>using namespace std;typedef long long LL;typedef vector<LL> vec;typedef vector<vec> mat;const int maxn = 60;LL mod = 1 << 16;LL t, n, k;LL yh[maxn][maxn];int cas = 0;void get_yh(){    yh[0][0] = yh[1][0] = yh[1][1] = 1;    for(int i= 2; i<= 50; i++)    {        yh[i][0] = 1;        for(int j= 1; j<= i; j++)            yh[i][j] = (yh[i - 1][j - 1] + yh[i - 1][j]);    }}mat mul(mat A, mat B){    mat C(A.size(), vec(B[0].size()));    for(int i= 0; i< A.size(); i++)        for(int k= 0; k< B.size(); k++)            if(A[i][k] != 0)            for(int j= 0; j< B[0].size(); j++)                C[i][j] = (C[i][j] + A[i][k] * B[k][j]) % mod;    return C;}mat pow(mat A, LL n){    mat B(A.size(), vec(A.size()));    for(int i= 0; i< A.size(); i++)        B[i][i] = 1;    while(n > 0)    {        if(n & 1) B = mul(B, A);        A = mul(A, A);        n >>= 1;    }    return B;}void solve(){    mat B(k+2, vec(1));    for(int i= 0; i< k+2; i++)        B[i][0] = 1;    mat A(k+2, vec(k+2));    A[0][0] = 1;    for(int i= 1; i< k+2; i++)        A[0][i] = yh[k][i - 1];    for(int i= 1; i< k+2; i++)    {        for(int j= 0; j<= k - i + 1; j++)            A[i][i+j] = yh[k - i + 1][j];    }    A = pow(A, n - 1);    B = mul(A, B);    printf("Case %d: %lld\n", ++cas, B[0][0]);}int main(){    //freopen("in.txt", "r", stdin);    mod *= mod;    get_yh();    scanf("%lld", &t);    int cas = 0;    while(t --)    {        scanf("%lld %lld", &n, &k);        solve();    }    return 0;}