POJ 3100 & ZOJ 2818 & HDU 2740 Root of the Problem(数学)

题目链接：

POJ:

id=3100" style="font-size:18px">http://poj.org/problem?

id=3100

ZOJ:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=1818

HDU:

pid=2740">http://acm.hdu.edu.cn/showproblem.php?pid=2740

Description

Given positive integers B and N, find an integer A such that A^N is as close as possible to B. (The result A is an approximation to the Nth root of B.) Note that A^N may be less than, equal to, or greater than B.

Input

The input consists of one or more pairs of values for B and N. Each pair appears on a single line, delimited by a single space. A line specifying the value zero for both B and N marks the end of the input. The value of B will be in the range 1 to 1,000,000
(inclusive), and the value of N will be in the range 1 to 9 (inclusive).

Output

For each pair B and N in the input, output A as defined above on a line by itself.

Sample Input

4 3
5 3
27 3
750 5
1000 5
2000 5
3000 5
1000000 5
0 0

Sample Output

1
2
3
4
4
4
5
16

Source

Mid-Central USA 2006

题意：

给定整数b和n，求整数a使得a^n最接近b。

代码例如以下：

#include <cstdio>
#include <cmath>
int main()
{
    int b, n;
    while(~scanf("%d%d",&b,&n))
    {
        if(b==0 && n==0)
            break;
        double tt = (double)pow(b,1.0/n);
        int t1 = ceil(tt);//向上取整
        int t2 = floor(tt);//向下取整
        if(b-pow(t2,n)>pow(t1,n)-b)
            printf("%d\n",t1);
        else
            printf("%d\n",t2);
    }
    return 0;
}