HDU-1061 Rightmost Digit (快速幂的应用)

HDU-1061 Rightmost Digit (快速幂的应用)

Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

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.

 

Author

Ignatius.L

 

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技巧 任意找一个小的取模的数（肯定是要大于想要取模的位数）将次方值降下来

#include<stdio.h>
long long p=1e3;
long long pow_mod(long long a,long long x)
{
	long long ans=1;               //p是取模的数
    long long base=a%p;
	while(x>0)
	{
       if((x&1)==1)
       ans=(ans*base)%p;     //if里的条件也可以写成x&1
	   base=base*base%p;
	   x=x>>1;
	}
	return ans;
}
int main()
{
    int t;
    long long n,ans;
    while(~scanf("%lld",&t))
    {
        while(t--)
        {
            scanf("%lld",&n);
            ans=pow_mod(n,n)%10;
            printf("%lld\n",ans);
        }
    }
    return 0;
}