本文主要是介绍poj 3070 矩阵快速幂，希望对大家解决编程问题提供一定的参考价值，需要的开发者们随着小编来一起学习吧！
| Time Limit: 1000MS | Memory Limit: 65536K | |
| Total Submissions: 13073 | Accepted: 9295 |
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:
题意求数列的n项
矩阵快速幂其实就是在幂那里二分优化，和快速乘方是一样的意思
一直以为还有什么更好的方法让三层循环变少，天真了一段时间
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;#define LL long long
#define MOD 10000struct mat
{int m[3][3];
}ans,base;mat multiply(mat a,mat b)
{mat temp;for(int i=0;i<2;i++)for(int j=0;j<2;j++){temp.m[i][j]=0;for(int k=0;k<2;k++)temp.m[i][j]=(temp.m[i][j]+a.m[i][k]*b.m[k][j])%MOD;}return temp;
}int quickmod(LL n)
{base.m[0][0]=base.m[0][1]=base.m[1][0]=1;base.m[1][1]=0;ans.m[0][0]=ans.m[1][1]=1;ans.m[1][0]=ans.m[0][1]=0;while(n){if(n&1)ans=multiply(ans,base);base=multiply(base,base);n>>=1;}return ans.m[1][0];
}int main()
{LL n;//freopen("in.txt","r",stdin);while(scanf("%I64d",&n),n!=-1){printf("%d\n",quickmod(n));}return 0;
}
这篇关于poj 3070 矩阵快速幂的文章就介绍到这儿，希望我们推荐的文章对编程师们有所帮助！
