計算 isp while code single lan ostream 進制 代碼

題目鏈接 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. 代碼:

 1 #include<iostream>
 2 using namespace std;
 3 int f(int a,int b){
 4     int ans=1;
 5     a = a % 10;
 6     while(b > 0){
 7         if(b & 1)
 8             /**
 9                 1.b & 1取b二進制的最低位，判斷和1是否相同，相同返回1，否則返回0，可用於判斷奇偶
10                 2.b>>1//把b的二進制右移一位，即去掉其二進制位的最低位
 
11             */
12             ans = (ans * a) % 10;
13             b = b >> 1;
14             a = (a * a)%10;
15     }
16     return ans;
17 }
18 int main(){
19         int n,t;
20         int result;
21         cin>>t;
22         while(t--){
23             cin>>n;
24             result=f(n,n);//
 
計算n的b次方
25             cout<<result<<endl;
26         }
27         return 0;
28 }

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Rightmost Digit