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Fibonacci
Description In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn. 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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn 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
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图片来源：
http://www.cnblogs.com/xudong-bupt/archive/2013/03/19/2966954.html
因此斐波那契数列可用矩阵表示
#include<cstdio>
#include<cstdlib>
#include<cstring>
using namespace std;
const int MOD=10000;
int ans[2][2];
int a[2][2]={1,1,1,0};
void Mulit(int A[2][2],int B[2][2]){int D[2][2]={0};for(int i=0;i<2;++i){for(int k=0;k<2;++k){if(A[i][k]){for(int j=0;j<2;++j){D[i][j]=(D[i][j]+(A[i][k]*B[k][j]))%MOD;}}}}memcpy(A,D,sizeof(A)*4);
}
void Matrix(int A[2][2],int n){ans[0][0]=ans[1][1]=1;ans[0][1]=ans[1][0]=0;while(n){if(n&1)Mulit(ans,A);Mulit(A,A);n>>=1;}
}
int main()
{int n,i,j,k;while(scanf("%d",&n)&&n!=-1){a[0][0]=a[0][1]=a[1][0]=1;a[1][1]=0;Matrix(a,n);printf("%d\n",ans[0][1]);}return 0;
}