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=;

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

Matrix unit,init;

Matrix Mul(Matrix a,Matrix b){
    Matrix c;
    for(int i=;i<;i++)
        for(int j=;j<;j++){
            c.arr[i][j]=;
            for(int k=;k<;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&){
            b=Mul(b,a);
        }
        a=Mul(a,a);
        k>>=;
    }
    return b;
}

void Init(){
    unit.arr[][]=,   unit.arr[][]=,   unit.arr[][]=unit.arr[][]=;

    init.arr[][]=init.arr[][]=,    init.arr[][]=,   init.arr[][]=;
}

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-);
        int ans=(*res.arr[][]-)%mod;    //注意这里，刚开始这里没有%mod，WA了好几次
        printf("%d\n",ans);
    }
    return ;
}