【POJ】3070

Fibonacci
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 12951 Accepted: 9210
Description

In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 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 Fn.

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 Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).

Sample Input

0
9
999999999
1000000000
-1
Sample Output

0
34
626
6875
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

Stanford Local 2006
入门题目
代码

#include<bits/stdc++.h>#define LL long longusing namespace std;const int MAXN =100+10;const int mod=10000;struct Matrix{    LL a[MAXN][MAXN];    int r,c;};Matrix ori,res;void init(){    memset(res.a,0,sizeof(res.a));    res.r=2;res.c=2;    for(int i=1;i<=2;i++)  //将结果矩阵赋值为单位矩阵        res.a[i][i]=1;//单位矩阵A就是 [未知矩阵]B*A=B;    ori.r=ori.c=2;    ori.a[1][1]=ori.a[1][2]=ori.a[2][1]=1;// 初始化初始矩阵    ori.a[2][2]=0;}Matrix multi(Matrix x,Matrix y){    Matrix z;    memset(z.a,0,sizeof(z.a));    z.r=x.r;  z.c=y.c; //新矩阵的行数等于x的行数，列数等于y的列数    for(int i=1;i<=x.r;i++){        for(int k=1;k<=x.c;k++){            if(x.a[i][k]==0) continue;            for(int j=1;j<=y.c;j++)                z.a[i][j]=(z.a[i][j]+(x.a[i][k]*y.a[k][j]%mod))%mod;        }    }    return z;}void Matrix_mod(int n){    while(n){        if(n&1) res=multi(ori,res);        ori=multi(ori,ori);        n>>=1;    }    printf("%lld\n",res.a[1][2]%mod);}int main(){    int n;    while(scanf("%d",&n)&&(n!=-1)){        init();        Matrix_mod(n);    }    return 0;}