【P-oj】-Fibonacci（矩阵）

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 14466 Accepted: 10216

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

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:

题解：和第一道矩阵的题一样，只不过这里还有 F0=0而已；

#include <cstdio>#include <stack>#include <queue>#include <cmath>#include <vector>#include <cstring>#include <algorithm>using namespace std;#define CLR(a,b) memset(a,b,sizeof(a))#define INF 0x3f3f3f3f#define LL long longconst LL MOD = 1e4;struct Matrix{int h,w;int m[5][5];};Matrix Matrix_multiply(Matrix a,Matrix b){Matrix c;c.h=a.h;c.w=b.w;CLR(c.m,0);for(int i=1;i<=a.h;i++){for(int j=1;j<=a.w;j++){if(a.m[i][j]==0)continue;for(int k=1;k<=b.w;k++)c.m[i][k]=(c.m[i][k]+a.m[i][j]*b.m[j][k]%MOD)%MOD;}}return c;}Matrix Quick(Matrix a,int n){Matrix ans;ans.h=ans.w=a.h;CLR(ans.m,0);for(int i=1;i<=a.h;i++)ans.m[i][i]=1;while(n){if(n&1)ans=Matrix_multiply(ans,a);n>>=1;a=Matrix_multiply(a,a);}return ans;}int main(){int n;while(~scanf("%d",&n)){if(n==0)printf("0\n");else if(n==-1)break;else{Matrix pr,now,anss;pr.h=pr.w=2;CLR(pr.m,0);pr.m[1][1]=pr.m[1][2]=pr.m[2][1]=1,pr.m[2][2]=0;now.h=1,now.w=2;now.m[1][1]=1,now.m[1][2]=1;CLR(anss.m,0);anss=Matrix_multiply(now,Quick(pr,n-1));printf("%d\n",anss.m[1][2]);}}return 0;}