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矩阵乘法 POJ3070 Fibonacci

2024-10-22 03:11:40   211人阅读

Fibonacci
Time Limit: 1000MS   Memory Limit: 65536K
Total Submissions: 15215   Accepted: 10687

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

技术分享.

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

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Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:

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Source

Stanford Local 2006
 
 1 #include<iostream>
 2 #include<cstdio>
 3 #include<cstring>
 4 #include<algorithm>
 5 using namespace std;
 6 int n,a[2][2],b[2][2];
 7 void mul(int a[2][2],int b[2][2],int ans[2][2]){
 8     int t[2][2];
 9     for(int i=0;i<2;i++)
10         for(int j=0;j<2;j++){
11             t[i][j]=0;
12             for(int k=0;k<2;k++) t[i][j]=(t[i][j]+a[i][k]*b[k][j])%10000; 
13         }
14     for(int i=0;i<2;i++)
15         for(int j=0;j<2;j++) ans[i][j]=t[i][j];
16 } 
17 int main(){
18     while(scanf("%d",&n)){
19         if(n==-1) return 0;
20         a[0][0]=a[1][0]=a[0][1]=b[0][0]=b[1][1]=1;
21         a[1][1]=b[1][0]=b[0][1]=0;
22         while(n){
23             if(n&1) mul(a,b,b);
24             n>>=1;
25             mul(a,a,a);
26         }
27         printf("%d\n",b[1][0]);
28     }
29     return 0;
30 }

 

矩阵乘法 POJ3070 Fibonacci


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