POJ3070 Fibonacci

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 13731		Accepted: 9727

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, and F_n = F_{n − 1} + F_{n − 2} for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:

0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …

An alternative formula for the Fibonacci sequence is

.

Given an integer n, your goal is to compute the last 4 digits of F_n.

Input

The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.

Output

For each test case, print the last four digits of F_n. If the last four digits of F_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print F_n mod 10000).

Sample Input

0
9
999999999
1000000000
-1

Sample Output

0
34
626
6875

Hint

As a reminder, matrix multiplication is associative, and the product of two 2 × 2 matrices is given by

.

Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

.

Source

Stanford Local 2006

矩阵乘法入门题。

hint已经写明了解法

/*by SilverN*/
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<vector>
using namespace std;
const int mod=1e4;
const int mxn=240;
struct mat{
    int s[3][3];
    mat(){memset(s,0,sizeof s);}
};
mat operator * (const mat a,const mat b){
    mat c;
    for(int i=1;i<=2;i++)
     for(int j=1;j<=2;j++)
       for(int k=1;k<=2;k++){
        (c.s[i][j]+=a.s[i][k]*b.s[k][j])%=mod;
       }
    return c;
}
int n;
int main(){
    int i,j;
    while(scanf("%d",&n) && n!=-1){
        mat a,b;
        a.s[1][1]=1;a.s[1][2]=1;a.s[2][1]=1;a.s[2][2]=0;
        b.s[1][1]=1;b.s[1][2]=0;b.s[2][1]=0;b.s[2][2]=1;
        for(;n;n>>=1,a=a*a)
            if(n&1)b=b*a;
        printf("%d
",b.s[2][1]);
    }
    return 0;
}