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
