HDU 2256 Problem of Precision （矩阵乘法）

Problem of Precision

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 686    Accepted Submission(s): 386

Problem Description

 

Input

The first line of input gives the number of cases, T. T test cases follow, each on a separate line. Each test case contains one positive integer n. (1 <= n <= 10^9)

 

Output

For each input case, you should output the answer in one line.

 

Sample Input

3 1 2 5

 

Sample Output

9 97 841

 

Source

HDOJ 2008 Summer Exercise（4）- Buffet Dinner

 

Recommend

lcy

 

 

 

 

 

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;

const int mod=1024;

struct Matrix{
    int arr[2][2];
};

Matrix unit,init;

Matrix Mul(Matrix a,Matrix b){
    Matrix c;
    for(int i=0;i<2;i++)
        for(int j=0;j<2;j++){
            c.arr[i][j]=0;
            for(int k=0;k<2;k++)
                c.arr[i][j]=(c.arr[i][j]+a.arr[i][k]*b.arr[k][j]%mod)%mod;
            c.arr[i][j]%=mod;
        }
    return c;
}

Matrix Pow(Matrix a,Matrix b,int k){
    while(k){
        if(k&1){
            b=Mul(b,a);
        }
        a=Mul(a,a);
        k>>=1;
    }
    return b;
}

void Init(){
    unit.arr[0][0]=5,   unit.arr[0][1]=2,   unit.arr[1][0]=unit.arr[1][1]=0;

    init.arr[0][0]=init.arr[1][1]=5,    init.arr[0][1]=2,   init.arr[1][0]=12;
}

int main(){

    //freopen("input.txt","r",stdin);

    int t,n;
    Init();
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        Matrix res=Pow(init,unit,n-1);
        int ans=(2*res.arr[0][0]-1)%mod;    //注意这里，刚开始这里没有%mod，WA了好几次
        printf("%d
",ans);
    }
    return 0;
}