poj3070(快速幂)

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7250 Accepted: 5140

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

矩阵的快速幂，套用模板

#include <cstdlib>#include <cstring>#include <cstdio>#include <iostream> using namespace std;int N;struct matrix{int a[3][3];}origin,res;matrix multiply(matrix x,matrix y){matrix temp;memset(temp.a,0,sizeof(temp.a));for(int i=0;i<2;i++){for(int j=0;j<2;j++){for(int k=0;k<2;k++){temp.a[i][j]=(temp.a[i][j]+x.a[i][k]*y.a[k][j])%mod;}}}return temp;}void calc(int n){int i,j;while(n){if(n&1)res=multiply(res,origin);n>>=1;origin=multiply(origin,origin);}}int main(){    while(cin>>N&&N>=0)    {res.a[0][0]=res.a[1][1]=1;res.a[0][1]=res.a[1][0]=0;origin.a[0][0]=origin.a[0][1]=origin.a[1][0]=1;origin.a[1][1]=0;        calc(N);printf("%d\n",res.a[0][1]);    }    return 0;}