1061 快速幂取模

 HDU 1061 Rightmost Digit

分类： ACM 算法 2011-12-17 17:37 749人阅读 评论(2) 收藏 举报

integeroutputinputeach算法c

Problem Description

Given a positive integer N, you should output the most right digit of N^N.
Input

The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.

Each test case contains a single positive integer N(1<=N<=1,000,000,000).
Output

For each test case, you should output the rightmost digit of N^N.
Sample Input

2

3

4
Sample Output

7

6
Hint
In the first case, 3 * 3 * 3 = 27, so the rightmost digit is 7.

In the second case, 4 * 4 * 4 * 4 = 256, so the rightmost digit is 6.
题意:

求n^n模10

代码:

[cpp] view plaincopy

#include<stdio.h>
int exp_mod(int a,unsigned int n,int b)

{

    int t;

    if(n==0||n==1)

    return n==0?1:a;

    t=exp_mod(a,n/2,b);

    t=t*t;

    if(n%2==1)t=t*a;

    return t%b;

}

int main()

{

    int t;

    scanf("%d",&t);

    while(t--)

    {

        long long n;

        scanf("%lld",&n);

        printf("%d\n",exp_mod(n%10,n,10));

    }

    return 0;

}
分析:

代码是在网上找的模版,第一次做是想找规律的可是实在是太繁琐勒(其实是我懒= =! ),就上网看看有没有更简单的方法,才知道这样的题有模版,就是快速幂取模.果断背下来....

下面是我经过在网上找的解析:

快速幂取模就是在O(logn)内求出a^n mod b的值。算法的原理是ab mod c=(a mod c)(b mod c)mod c

1061 快速幂取模