hdu 1061Rightmost Digit 模幂运算

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.

n^n的最后一位
模幂运算
快速幂
···c++

include

include

include

include

include

include

include

include

include

pragma comment(linker,”/STACK:102400000,102400000”)

define pi acos(-1.0)

define EPS 1e-6

define INF (1<<24)

using namespace std;
int pow2(int a,int b)
{
int r=1,base=a;
while(b!=0)
{
if(b%2)
r=r*base%10;
base=base*base%10;
b/=2;
}
return r;
}
int main()
{
int a,b;
int T;
scanf(“%d”,&T);
while(T–)
{
scanf(“%d”,&a);
//a=a%10;
printf(“%d\n”,pow2(a%10,a));
}
return 0;
}
···