LightOJ - 1132 Summing up Powers

Given N and K, you have to find

(1^K + 2^K + 3^K + ... + N^K) % 2³²

Input

Input starts with an integer T (≤ 200), denoting the number of test cases.

Each case contains two integers N (1 ≤ N ≤ 10¹⁵) and K (0 ≤ K ≤ 50) in a single line.

Output

For each case, print the case number and the result.

Sample Input

3

3 1

4 2

3 3

Sample Output

Case 1: 6

Case 2: 30

Case 3: 36

没什么好说的，普通的快速幂肯定会超时，所以这里就用到矩阵快速幂。

这个问题最重要的是推出状态转移方程。

把操作矩阵D表示出来即可

#include<iostream>
#include<stdio.h>
#include<string>
#include<math.h>
#include<vector>
#include<string.h>
#include<iterator>
using namespace std;
typedef unsigned long long ll;

ll n,k;
const ll mod = ((ll)1<<32);

struct Matrix{
    ll a[55][55];
    void init(){
        for(int i = 0; i< k+ 2 ;i++){
            for(int j = 0; j< k + 2; j ++){
                if(i==j)
                    a[i][j] = 1;
                else
                    a[i][j] = 0;
            }
        }
    }
    void init2(){
        for(int i = 0; i< k+ 2 ;i++){
            for(int j = 0; j< k + 2; j ++){
                a[i][j] = 0;
            }
        }
    }
}A;//矩阵表示


void init1(){
    int i,j;
    A.a[0][0] = 1;
    for(i=1;i<k+2;i++){
        for(j=0;j<i;j++){
            A.a[i][j] = 0;
        }
    }
    for(i=0;i<k+2;i++)
        A.a[i][k+1] = 1;
    for(i=k;i>0;i--){
        for(j=k;j>=0;j--){
            A.a[i][j] = A.a[i+1][j] + A.a[i+1][j+1];
        }
    }
    for(j=1;j<k+2;j++)
        A.a[0][j] = A.a[1][j];
    /*for(i=0;i<k+2;i++){
        for(j=0;j<k+2;j++){
            cout<<A.a[i][j]<<" ";
        }
        cout<<endl;
    }*/
}

Matrix multiply(Matrix x,Matrix y){
    Matrix ans;
    ans.init2();
    for(int i = 0 ;i < k+2; i++){
        for(int p = 0; p < k+2 ;p++)
        for(int j = 0; j < k+2; j++){//改变p和j的顺序能有效的改进复杂度
                ans.a[i][j]=(ans.a[i][j]+(x.a[i][p]*y.a[p][j])%mod)%mod;//注意取模的地方
            }
        }

    return ans;
}//矩阵的乘法

Matrix Matrix_pow(Matrix ma,ll n){
    Matrix ans;
    ans.init();

    while(n){
        if(n%2==1)
            ans = multiply(ans,ma);

        ma = multiply(ma,ma);
        n>>=1;
    }
    return ans;
}//矩阵的快速幂

int main()
{
    int t;
    int i,j;
    ll result;
    int a,b;
    scanf("%d",&t);

    for(j=1;j<=t;j++){
        scanf("%lld %lld",&n,&k);
        init1();
        result = 0;
        Matrix x = Matrix_pow(A,n-1);

        /*for(a = 0 ;a<k+2;a++){
            for(b=0;b<k+2;b++){
                cout<<x.a[a][b]<<" ";
            }
            cout<<endl;
        }*/

        for(i=0;i<k+2;i++){
            result = (result + x.a[0][i]%mod)%mod;
        }
        cout<<"Case "<<j<<": "<<result<<endl;
    }
    return 0;
}