数学函数

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.

<span style="font-size:14px;">#include<stdio.h>
#include<math.h>
int main()
{
        int b,n;
        while(scanf("%d %d",&b,&n),b,n)
        {
                int i, temp1, temp2;
                for(i=1;i<=b;i++)
                {
                        if(int(pow((float)i,(int)n))==b)<span style="color:#33CC00;">//注意b是整形，所以要进行强制转换</span>
                        {
                                printf("%d\n",i);
                                break;
                        }
                        else if(pow((float)i,(int)n)<b&&pow((float)i+1,(int)n)>b)
                        {
                        	temp1=pow((float)i,(int)n);
                        	temp2=pow((float)i+1,(int)n);
                        	if(temp1-b>0&&temp2-b>0)
                        	{
	                        	if(temp1-b<temp2-b)
	                        	{
	                        		printf("%d\n", i);
	                        		break;
	                        	}
	                        	else
	                        	{
	                        		printf("%d\n", i+1);
	                        		break;
	                        	}
	                        }
	                        else if(b-temp1>0&&b-temp2>0)
	                        {
                        		if(b-temp1<b-temp2)
                        		{
		                        	printf("%d\n", i);
	                        		break;
		                        }
		                        else
	                        	{
	                        		printf("%d\n", i+1);
	                        		break;
	                        	}
                        	}
                       	 	else if(b-temp1>0&&temp2-b>0)
	                        {
                        		if(b-temp1<temp2-b)
                        		{
		                        	printf("%d\n", i);
	                        		break;
		                        }
		                        else
	                        	{
	                        		printf("%d\n", i+1);
	                        		break;
	                        	}
                        	}
                        	else if(temp1-b>0&&b-temp2>0)
                        	{
	                        	if(temp1-b<b-temp2)
                        		{
		                        	printf("%d\n", i);
	                        		break;
		                        }
		                        else
	                        	{
	                        		printf("%d\n", i+1);
	                        		break;
	                        	}
	                        }
                        }
                }
        }
        return 0;
}</span>