poj3100

http://poj.org/problem?id=3100

Root of theProblem

Time Limit: 1000MSMemory Limit: 65536KTotal Submissions: 10385Accepted: 5560

Description

Given positive integers B and N, find an integer A such thatA^N is as close as possible to B. (The result A is anapproximation to the Nth root of B.) Note that A^N may beless 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. Aline specifying the value zero for both B and N marks the end ofthe 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 ona line by itself.

Sample Input

4 35 327 3750 51000 52000 53000 51000000 50 0

Sample Output

123444516

Source

Mid-Central USA 2006

题意给你两个数b,n求一个数a使得a的n次方最接近b。。。水过。。pow()用的时候要注意格式啊

在math.h中，函数pow有三种重载形式：
long double pow(long double,int)
float pow(float,int)
double pow(double,int)
对于所给的参数int,int，编译器无法判断应该匹配哪个函数，因此报错
可以将代码改为pow(10.0,(int)i)

#include<stdio.h>#include<math.h>int main(){        int b,n;        while(scanf("%d %d",&b,&n),b,n)        {                int i;                for(i=1;i<=b;i++)                {                        if(pow((float)i,(int)n)==b)                        {                                printf("%d\n",i);                                break;                        }                        else if(pow((float)i,(int)n)<b&&pow((float)i+1,(int)n)>b)                        {                                if(abs(pow((float)i,(int)n)-b)<abs(pow((float)i+1,(int)n)-b))                                {                                        printf("%d\n",i);                                        break;                                }                                else                                {                                        printf("%d\n",i+1);                                        break;                                }                        }                }        }        return 0;}