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).
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
Recommend
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; }