poj 3070 Fibonacci

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 7149 Accepted: 5053

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

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:

.

题意：

求第n个fibonacci数后四位值

解析：

根据题目给定的方法，可以肯定会用到求快速幂的方法。快速幂就是当n为偶数时，a^n=a^(n/2)*a^(n/2),而后递归求;当n为奇数时,a^n=a^((n-1)/2)*a^((n-1)/2)*a，而后往下递归a^((n-1)/2),此处不同的就是矩阵的乘法，可以通过结构来定义

#include<stdio.h>struct matrix{int v[2][2];};//为了递归的时候返回值方便，所以把矩阵定义在结构里面matrix num;matrix multi(matrix a,matrix b)//矩阵a乘矩阵b{matrix m;m.v[0][0] =m.v[0][1] = m.v[1][0]= m.v[1][1] = 0;int i,j,k;for(i = 0 ; i < 2; i ++)for(j = 0 ; j < 2; j ++)for(k = 0; k < 2; k ++)m.v[i][j] = (m.v[i][j] + a.v[i][k] * b.v[k][j] ) %10000;//此处不可以换成m.v[i][j] += (a.v[i][k] * b.v[k][j] ) %10000;，因为这是会叠加超过10000的，//例如m.v[i][j]等于9000时，a.v[i][k] * b.v[k][j] 等于3000，加一起就超过10000了return m; }matrix fx(int n)//num的n次方{matrix t,a;if(n == 1)return num;if(n%2==0){a = fx(n/2);t = multi(a,a);}else{a = fx((n-1)/2);t = multi( a,a );t = multi(t,num);}return t;}int main(){int n;while(scanf("%d",&n) != EOF && n != -1) {int i;num.v[0][0] = num.v[0][1] = num.v[1][0]= 1;num.v[1][1] = 0;if(n == 0)printf("0\n");elseprintf("%d\n",fx(n-1).v[0][0]);}return 0;}