Fibonacci

这次写的还不错，就有一个小错误

D - Fibonacci Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%lld & %llu Submit  Status

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., printF_n 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:

.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 #include <iostream> #include <cstdio> #include <cstring> #define mod 10000 using namespace std; struct matrix { int ma[3][3]; }; matrix mult(matrix a,matrix b) { matrix ans; memset(ans.ma,0,sizeof(ans.ma)); for(int i=1;i<=2;i++) { for(int j=1;j<=2;j++) { for(int k=1;k<=2;k++) { ans.ma[i][j]+=(a.ma[i][k]*b.ma[k][j])%mod; } ans.ma[i][j]%=mod; } } return ans; } matrix pow(matrix x,int n) { matrix ans; memset(ans.ma,0,sizeof(ans.ma)); for(int i=1;i<=2;i++) //此处注意变化行列数 { ans.ma[i][i]=1; } while(n) { if(n&1) { ans=mult(ans,x); } x=mult(x,x); n>>=1; } return ans; } int main() { int n; while(~scanf("%d",&n)) { if(n==-1) break; if(n==0) { printf("0\n"); continue; } matrix sum; sum.ma[1][1]=sum.ma[1][2]=sum.ma[2][1]=1; sum.ma[2][2]=0; sum=pow(sum,n); printf("%d\n",sum.ma[1][2]); } return 0; }