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Rightmost Digit
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 6667 Accepted Submission(s): 1721 |
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 234 |
Sample Output 76 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 |
思路： long long int 因为已知容易超出限制 ，所以此处一定要注意 ,另外，利用/2的方式进行 这样可以大大缩短程序运行时间。
1 #include<stdio.h> 2 //大数问题 3 long long Solve(long long n) 4 { 5 long long int count=n; 6 long long int result=1; 7 while(count) 8 { 9 if(count%2!=0)10 {11 result*=n;12 if(result>=10)13 result=result%10;14 }15 n*=n;16 n=n%10; 17 count/=2;18 }19 return result;20 }21 22 23 int main(int argc, char *argv[])24 {25 long long T,N;26 scanf("%lld",&T);27 while(T--)28 {29 scanf("%lld",&N);30 printf("%lld\n",Solve(N));31 32 } 33 return 0;34 }
Rightmost Digit
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