【poj 3070】Fibonacci

D - Fibonacci

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

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:

.

Source

代码：

#include<cstdio>#include<cstring>#include<algorithm>#define LL long long#define Mod 10000#define CLR(a,b) memset(a,b,sizeof(a))using namespace std;struct Mat{int a[4][4];int h,w;}pr,ans;void init(){ans.a[2][1]=ans.a[1][2]=0;  //初始化过渡矩阵 ans.h=ans.w=2;for(int i=1;i<=2;i++)ans.a[i][i]=1;pr.a[1][1]=pr.a[1][2]=pr.a[2][1]=1; //初始化单位矩阵 pr.a[2][2]=0;pr.h=pr.w=2;}Mat Mat_Mul(Mat x,Mat y)  //矩阵相乘 {Mat t;CLR(t.a,0);t.h=x.h;t.w=y.w;for(int i=1;i<=x.h;i++){for(int j=1;j<=x.w;j++){if(x.a[i][j]==0) continue;for(int k=1;k<=y.w;k++)t.a[i][k]=(t.a[i][k]+x.a[i][j]*y.a[j][k]%Mod)%Mod;}}return t;}void Mat_mod(int n)  //矩阵快速幂 {while(n){if(n&1)ans=Mat_Mul(pr,ans);pr=Mat_Mul(pr,pr);n>>=1;}}int main(){    int n;    while(~scanf("%d",&n)&&n!=-1)    {    init();    Mat_mod(n);    printf("%d\n",ans.a[1][2]%Mod);}return 0;}