POJ 3070 Fibonacci （矩阵快速幂求fibonacci）

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 11439 Accepted: 8134

Description

In the Fibonacci integer sequence, F₀ = 0, F₁ = 1, andF_n = F_{n − 1} + F_{n − 2} forn ≥ 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 ofF_n are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., printF_n mod 10000).

Sample Input

099999999991000000000-1

Sample Output

0346266875

ac代码：

#include<stdio.h>#include<string.h>#include<math.h>#include<iostream>#include<algorithm>#define MAXN 1000000000#define MOD 10000#define LL long long#define INF 0xfffffff#define fab(a)(a)>0?(a):(-a)using namespace std; int fib(int b){int t[2][2];int ans[2][2];int a[2][2];ans[0][0]=ans[1][1]=1,ans[0][1]=ans[1][0]=0;a[0][0]=a[0][1]=a[1][0]=1,a[1][1]=0;int i,j,k;while(b){if(b%2){for(i=0;i<2;i++)for(j=0;j<2;j++)t[i][j]=ans[i][j];ans[0][0]=ans[1][1]=ans[0][1]=ans[1][0]=0;//每次使用完都要清零  因为是累加for(i=0;i<2;i++){for(j=0;j<2;j++){for(k=0;k<2;k++){ans[i][j]=(ans[i][j]+(t[i][k]*a[k][j]))%MOD;}}}}for(i=0;i<2;i++)for(j=0;j<2;j++)t[i][j]=a[i][j];a[0][0]=a[1][1]=a[0][1]=a[1][0]=0;//每次使用完都要清零  因为是累加for(i=0;i<2;i++){for(j=0;j<2;j++){for(k=0;k<2;k++){a[i][j]=(a[i][j]+(t[i][k]*t[k][j]))%MOD;}}}b/=2;}return ans[0][1];}int main(){int n;while(scanf("%d",&n)!=EOF){if(n==-1)break;printf("%d\n",fib(n));}    return 0;}