杭电 HDU 1060 Leftmost Digit
2015-03-23 16:32
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Leftmost Digit
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 14198 Accepted Submission(s): 5437
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).
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 #include<iostream> #include<cmath> using namespace std; int main() { int n;int T; cin>>T; while(T--) { scanf("%d",&n); double sum1=n*log10(double(n)); double sum2=sum1-floor(sum1); printf("%d\n",int (pow(10,sum2))); } return 0; }
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