HDOJ-1061 Rightmost Digit

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Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 47684 Accepted Submission(s): 18044

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.

题意是求n的n次方，并且仅输出最后一位。
数据比较大，在快速幂过程中对数据取10的余数即可。
代码如下：

#include <cstdio>#include <algorithm>#include<math.h>#include<string.h>using namespace std;__int64 quickpow(__int64 n,__int64 m,__int64 mod){    __int64 a=1,b=n;    while(m)    {        if(m&1) a=(a*b)%mod;        b=(b*b)%mod;        m>>=1;    }    return a;}int main(){    int t;    scanf("%d",&t);    while(t--)    {        __int64 n,mod=10;        scanf("%I64d",&n);        printf("%I64d\n",quickpow(n,n,mod));    }    return 0;}

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