Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12229 Accepted Submission(s): 4674
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1060
Problem Description
Given a positive integer N, you should output the leftmost 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 leftmost digit of N^N.
Sample Input
2
3
4
Sample Output
2
2
Hint
In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2.
In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.
Author
Ignatius.L
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代码:
1 #include<iostream> 2 #include<cmath> 3 using namespace std; 4 int main() 5 { 6 int sum; 7 while(cin>>sum) 8 { 9 while(sum--) 10 { 11 double n; 12 scanf("%lf",&n); 13 double x=n*log10(n*1.0); 14 _int64 y=(_int64)x; 15 double xy=x-y; 16 int temp=(int)pow(10.0,xy); 17 printf("%d ",temp); 18 } 19 } 20 return 0; 21 }