POJ 3100:Root of the Problem

Root of the Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 12060		Accepted: 6469

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

你  离  开  了  ，  我  的  世  界  里  只  剩  下  雨  。  。  。

#include <cstdio>
#include <cstdlib>
#include <cmath>
int main()
{
    int min, tmp;
    int n, b;
    float right;
    int low, high;
    while (scanf("%d%d", &b, &n), n != 0 || b != 0)
    {
        if (n == 1 || b == 1)
        {
            printf("%d
", b);
            continue;
        }
        right = pow(b, 1.0 / n);
        low = (int)right;
        high = (int)(right + 0.9999);
        min = b - pow(low, n);
        tmp = pow(high, n) - b;
        if (tmp < min)printf("%d
", high);
        else printf("%d
", low);
    }
    return 0;
}