Rightmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 36563 Accepted Submission(s): 13871

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 求N的N次方的最后一位数。快速幂。。假如8^4。可以看成64^2 。。这样运算次数减少了三次。假如8^5 我们可以先将一个8存起来，变成8^4。。然后一直反复操作。。代码

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              using namespace std; #define inf 0x6f6f6f6f #define mod 10 long long wei(long long k) { long long n=k; long long r=1; while(k) { if(k&1) //k次幂为奇数，可以存起一个数。 r=r*n%mod; n=n*n%mod; k>>=1; //位运算，加快速度。 } return r; } int main() { long long k; int t; scanf("%d",&t); while(t--) { cin>>k; long long ans=wei(k); cout<

HDU 1061 Rightmost Digit

Rightmost Digit