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| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 11024 | Accepted: 7846 |
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
#include<iostream> #include<cstdio> using namespace std; int n,a[2][2],b[2][2]; void mul(int a[2][2],int b[2][2],int ans[2][2]) { int t[2][2]; for (int i=0;i<=1;i++) for (int j=0;j<=1;j++) { t[i][j]=0; for (int k=0;k<=1;k++) t[i][j]=(t[i][j]+a[i][k]*b[k][j])%10000; } for (int i=0;i<=1;i++) for (int j=0;j<=1;j++) ans[i][j]=t[i][j]; } int main() { while (scanf("%d",&n)) { if (n==-1) return 0; a[0][0]=a[0][1]=a[1][0]=1; a[1][1]=0; b[0][0]=b[1][1]=1; b[0][1]=b[1][0]=0; while (n) { if (n&1) mul(a,b,b); n>>=1; mul(a,a,a); } printf("%d\n",b[1][0]); } }
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原文地址：http://www.cnblogs.com/ws-fqk/p/4737476.html
