【快速幂】——hdu1061
Rightmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 46398 Accepted Submission(s): 17500
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.法一:快速幂取模
#include <iostream> using namespace std; int qmod(long long a,long long b,long long c) { int ans = 1; a%=c; while(b>0) { if(b&1) ans = (ans * a) % c; b = b>>1; a = (a * a) % c; } return ans; } int main() { long long n,t; cin>>n; while(n--) { cin>>t;; cout<<qmod(t,t,10)<<endl; } return 0; }
法二:找规律
#include<stdio.h> int main() { __int64 i,n,s,m; while(scanf("%I64d",&n)!=EOF) { while(n--) { scanf("%I64d",&m); s=m; for (i=2;i<=m%4+4;i++) { s=s*m%10; } printf("%I64d\n",s); } } return 0; }