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.
.
题解：
按题目要求很容易写出矩阵
F[n] 0 1 1
F[n-1] 0 1 0
直接上矩阵快速幂即可
1 #include <algorithm> 2 #include <iostream> 3 #include <cstdlib> 4 #include <cstring> 5 #include <cstdio> 6 #include <cmath> 7 using namespace std; 8 typedef long long ll; 9 const int mod=10000; 10 int n; 11 struct matrix 12 { 13 ll a[3][3]; 14 matrix(){for(int i=1;i<=2;i++)for(int j=1;j<=2;j++)a[i][j]=0;} 15 matrix(ll b[3][3]){for(int i=1;i<=2;i++)for(int j=1;j<=2;j++)a[i][j]=b[i][j];} 16 inline matrix operator *(matrix p){ 17 matrix tmp; 18 for(int i=1;i<=2;i++) 19 for(int j=1;j<=2;j++){ 20 tmp.a[i][j]=0; 21 for(int k=1;k<=2;k++) 22 tmp.a[i][j]+=a[i][k]*p.a[k][j],tmp.a[i][j]%=mod; 23 } 24 return tmp; 25 } 26 }; 27 ll work(){ 28 if(n==0)return 0; 29 if(n==1||n==2)return 1; 30 n-=2; 31 ll t[3][3]={{0,0,0},{0,1,1},{0,0,0}};ll sum[3][3]={{0,0,0},{0,1,1},{0,1,0}}; 32 matrix S=matrix(t),T=matrix(sum); 33 while(n){ 34 if(n&1)S=S*T; 35 T=T*T;n>>=1; 36 } 37 return S.a[1][1]; 38 } 39 int main() 40 { 41 while(scanf("%d",&n)) 42 { 43 if(n==-1)break; 44 printf("%lld ",work()); 45 } 46 return 0; 47 }