hdu1061 Rightmost Digit 快速幂的简单应用
**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的n次方最右边的数,也就是最后一个数字。
思路:由于数值n的范围很大,不能用寻常套路写这道题,这道题的正解是快速幂取模运算的应用,基于公式:
(a*b)%c=(a%c*b%c)%c
(话说第一次WA就是因为取模出错,导致范围超了)
#include<stdio.h>
int f(int m,int n)
{
if( n == 1)
return m;
long long s = f(m,n/2)%10;
if( n%2 == 1)
return (s*s*m)%10;
else
return (s*s)%10;
}
int main()
{
int n;
int t;
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
printf("%d\n",f(n,n));
}
return 0;
}