Root of the Problem-3100

Root of the Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11955		Accepted: 6404

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

/***********************************************************************
    Copyright (c) 2015,wangzp
    All rights no reserved.
  
    Name: 《Root of the Problem》In PEKING UNIVERSITY ACM
    ID:   PROBLEM 3100
    问题简述: A的N次方尽量的接近B（其中，A,N,B都为正整数）
              根据输入的N和B，找出A
    Date: Aug 12, 2015 
 
***********************************************************************/
#include 
#include 

int main(void)
{
	long a,b,n;
	long c,tmp1,tmp2;
	/*注意以下输入终止的写法*/
	while(scanf("%ld %ld",&b,&n),b != 0 && n != 0)
	{
		c =(long)pow(b,1.0 / n);
		tmp1 = b - pow((double)c,n);/*重载需要指明参数确定类型*/
		tmp2 = pow((double)(c + 1),n) - b;
		if (tmp1 < tmp2)
		{
			printf("%d\n",c);
		} 
		else
		{
			printf("%d\n",c + 1);
		}
	}
	return 0;
}