【hdu 1060】【求N^N最低位数字】

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

[align=left]Problem Description[/align]
Given a positive integer N, you should output the leftmost digit of N^N.

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case, you should output the leftmost digit of N^N.

[align=left]Sample Input[/align]

2
3
4

[align=left]Sample Output[/align]

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.

[align=left]Author[/align]
Ignatius.L

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代码：

#include<iostream>
#include<cmath>
using namespace std;
int main()
{
int sum;
while(cin>>sum)
{
while(sum--)
{
double n;
scanf("%lf",&n);
double x=n*log10(n*1.0);
_int64 y=(_int64)x;
double xy=x-y;
int temp=(int)pow(10.0,xy);
printf("%d\n",temp);
}
}
return 0;
}

View Code