POJ3070 Fibonacci
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13731 | Accepted: 9727 |
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
矩阵乘法入门题。
hint已经写明了解法
1 /*by SilverN*/ 2 #include<algorithm> 3 #include<iostream> 4 #include<cstring> 5 #include<cstdio> 6 #include<cmath> 7 #include<vector> 8 using namespace std; 9 const int mod=1e4; 10 const int mxn=240; 11 struct mat{ 12 int s[3][3]; 13 mat(){memset(s,0,sizeof s);} 14 }; 15 mat operator * (const mat a,const mat b){ 16 mat c; 17 for(int i=1;i<=2;i++) 18 for(int j=1;j<=2;j++) 19 for(int k=1;k<=2;k++){ 20 (c.s[i][j]+=a.s[i][k]*b.s[k][j])%=mod; 21 } 22 return c; 23 } 24 int n; 25 int main(){ 26 int i,j; 27 while(scanf("%d",&n) && n!=-1){ 28 mat a,b; 29 a.s[1][1]=1;a.s[1][2]=1;a.s[2][1]=1;a.s[2][2]=0; 30 b.s[1][1]=1;b.s[1][2]=0;b.s[2][1]=0;b.s[2][2]=1; 31 for(;n;n>>=1,a=a*a) 32 if(n&1)b=b*a; 33 printf("%d ",b.s[2][1]); 34 } 35 return 0; 36 }