HDU1060 Leftmost Digit

1060

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18433 Accepted Submission(s): 7238

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

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m=n^n,兩邊分別對10取對數得 log10(m)=n*log10(n),得m=10^(n*log10(n))；

#include<stdio.h>
#include<math.h>
using namespace std;
int main() {
    int T,n;
    scanf("%d", &T);
    while (T--&&scanf("%d", &n) != EOF)
    {
        double temp1, temp2;
        temp1 = n*1.0*log10(1.0*n);
        temp2 = temp1 - (__int64)(temp1);
        printf("%.0lf\n", (floor)(pow(10.0, temp2)));
    }
    return 0;
}