Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 33161 Accepted Submission(s): 12696
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
题目大意:给你一个N,计算N^N个位上的数字是多少
思路:普通方法超时,利用高速幂取余计算N^N%10,这里贴一个二进制
高速幂取余的代码
#include#include __int64 QuickPow(__int64 a,__int64 p){ __int64 r = 1,base = a; __int64 m = 10; while(p!=0) { if(p & 1) r = r * base % m; base = base * base % m; p >>= 1; } return r;}int main(){ __int64 N; int T; scanf("%d",&T); while(T--) { scanf("%I64d",&N); __int64 ans = QuickPow(N,N); printf("%I64d\n",ans); } return 0;}