linux畅想

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 casesfollow.
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.

代码

#include<stdio.h>
#include<math.h>
int main()
{
    int n;
    double x,y;
    scanf("%d",&n);
    while(n--){
        scanf("%lf",&x);
        y=x*log10(x);
        y-=__int64(y);
        y=pow(10.0,y);
        printf("%I64d\n",__int64(y));
    }
    return 0;
}

分析
输入x，求x^x的最高位
因为存在整数y>0使得x^x=10^y;
又有y=y整数部分+y小数部分;
所以x^x=10^y整数部分*10^y小数部分;
因为“10^y整数部分”只有进位效果，对最高位的值没影响;
而1=<10^y小数部分<10,从而“10^y小数部分”的整数部分就是所求的;
应指出的是代码中两处用到了__int64而不用int是应为double型转int会溢出导致出错。