poj3070 Fibonacci fast power matrix template

Meaning of the questions:

Seeking Fibonacci of n items F [n] of Number film 10000 is f [n]% 10000
but 0 ≤ n ≤ 1,000,000,000
gave a formula:

analysis:

FIG matrix Fast Power n-1 th power of the matrix A 1 is three, the first row numbers in the first column is the answer.
(Remember modulo)
The following is the code board and ac

Quick copied to the matrix board power:

const int maxm=1e3+5;
struct Matrix{
    int a[maxm][maxm];
}res,ans;
Matrix mul(Matrix A,Matrix B,int n){//矩阵A，B和阶数n
    Matrix temp;//临时存放答案的矩阵
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            temp.a[i][j]=0;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            for(int k=1;k<=n;k++){
                temp.a[i][j]+=(A.a[i][k]%mod)*(B.a[k][j]%mod);
                temp.a[i][j]%=mod;
            }
        }
    }
    return temp;
}
void quickpower(int N,int n){//N是幂数，n是阶数
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            ans.a[i][j]=(i==j?1:0);
        }
    }
    while(N>0){
        if(N&1){
            ans=mul(ans,res,n);
        }
        res=mul(res,res,n);
        N>>=1;
    }
}

code：

#include<iostream>
#include<cstdio>
using namespace std;
const int maxm=3;
const int mod=1e4;
struct Matrix{
    int a[maxm][maxm];
}res,ans;
Matrix mul(Matrix A,Matrix B,int n){
    Matrix temp;
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            temp.a[i][j]=0;
        }
    }
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            for(int k=1;k<=n;k++){
                temp.a[i][j]+=(A.a[i][k]%mod)*(B.a[k][j]%mod);
                temp.a[i][j]%=mod;
            }
        }
    }
    return temp;
}
void quickpower(int N,int n){
    for(int i=1;i<=n;i++){
        for(int j=1;j<=n;j++){
            ans.a[i][j]=(i==j?1:0);
        }
    }
    while(N>0){
        if(N&1){
            ans=mul(ans,res,n);
        }
        res=mul(res,res,n);
        N>>=1;
    }
}
int main(){
    int n;
    while(cin>>n){
        if(n==-1)break;
        if(n==0){
            cout<<0<<endl;
            continue;
        }
        if(n==-1)break;
        res.a[1][1]=res.a[1][2]=res.a[2][1]=1;
        res.a[2][2]=0;
        quickpower(n-1,2);
        cout<<ans.a[1][1]<<endl;
    }
    return 0;
}