Time Limit: 1000MS | Memory Limit: 65536K | |
Total Submissions: 12329 | Accepted: 8748 |
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:
.
看了很多关于矩阵快速幂的题解,感觉矩阵快速幂得到的是一个含有多个项的矩阵,而答案只是其中一项,而且之还需要特判指数。入门的矩阵快速幂题。感受一下再写其它的题目。拿这题为例。题目中给出的项从0开始,F0=0,F1=1,F2=1,F3=2。。就是一个斐波那契的数列。由于用矩阵,那至少要两项来组成矩阵,根据他的递推公式。可以得到
[Fn,Fn-1]=[1*Fn-1+1*Fn-2,1*Fn-1+0*Fn-2]
代码:
#include<iostream> #include<algorithm> #include<cstdlib> #include<sstream> #include<cstring> #include<cstdio> #include<string> #include<deque> #include<stack> #include<cmath> #include<queue> #include<set> #include<map> using namespace std; typedef long long LL; #define INF 0x3f3f3f3f struct mat { LL m[2][2]; mat(){memset(m,0,sizeof(m));} }; mat cheng(mat a,mat b) { mat c; for (int i=0 ;i<2; i++) { for (int j=0; j<2; j++) { for (int k=0; k<2; k++) { c.m[i][j]+=(a.m[i][k]*b.m[k][j])%10000; } } } return c; } mat zxc(mat a,LL b) { mat c; c.m[0][0]=c.m[1][1]=1; while (b!=0) { if(b&1) c=cheng(c,a); a=cheng(a,a); b>>=1; } return c; } int main(void) { LL n; while (cin>>n&&n!=-1) { mat one; if(n==0) { cout<<0<<endl; continue; } else if(n==1) { cout<<1<<endl; continue; } one.m[0][0]=one.m[1][0]=one.m[0][1]=1; one=zxc(one,n-1); cout<<one.m[0][0]%10000<<endl;; } return 0; }