POJ 3070 Fibonacci 【矩阵快速幂取模 （模板）】

Fibonacci

Time Limit: 1000MS Memory Limit: 65536KTotal Submissions: 13262 Accepted: 9430

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:

题意：用矩阵快速幂取模求斐波拉契的项；

思路：直接上模板就行，在相乘的过程中进行取模运算；

失误：矩阵快速幂的代码有点麻烦，尽量写的简化一点；

AC代码：

#include<cstdio>#include<cstring>typedef long long LL;const LL MOD=10000;struct Matrix{LL Mat[3][3];};Matrix ori,res;Matrix Mat_mul(Matrix X,Matrix Y){Matrix Z; memset(Z.Mat,0,sizeof(Z.Mat));LL i=0,j=0,k=0;for(i=1;i<=2;++i){for(k=1;k<=2;++k){if(X.Mat[i][k]==0) continue;for(j=1;j<=2;++j){Z.Mat[i][j]+=(X.Mat[i][k]*Y.Mat[k][j])%MOD;Z.Mat[i][j]%=MOD;}}}return Z; }void Mat_Q(LL M){      ori.Mat[1][1]=1; ori.Mat[1][2]=1;     ori.Mat[2][1]=1; ori.Mat[2][2]=0;    res.Mat[1][1]=res.Mat[2][2]=1;res.Mat[1][2]=res.Mat[2][1]=0;    while(M){if(M&1) res=Mat_mul(res,ori);ori=Mat_mul(ori,ori);M>>=1;}}int main(){LL N;while(~scanf("%lld",&N),~N){    if(N==0) printf("0\n");else if(N==1) printf("1\n");else {         Mat_Q(N-1); printf("%lld\n",res.Mat[1][1]); }    } return 0; }