zoj2818 Root of the Problem 简单数学 开方

Root of the Problem

Time Limit:
2 Seconds Memory Limit:
65536 KB

Given positive integers B and N, find an integer A such that AN is as close as possible to B. (The result A is an approximation to the Nth root
of B.) Note that AN 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.

Example Input:	Example Output:
4 3 5 3 27 3 750 5 1000 5 2000 5 3000 5 1000000 5 0 0	1 2 3 4 4 4 5 16

Source: Mid-Central USA 2006





第一思路：二分找最小k，k^m<=n，然后和（k+1）^m比较。


第二思路：我们都知道n的m次方为pow（n，m）——精确值，则n开m次方为pow（n，1/m）——近似值（有时会把结果拿回去验证一遍）。当然这个函数适用于小数据，int范围就最好。

#include<cstdio>
#include<cstdlib>
#include<iostream>
#include<math.h>
#include<algorithm>
using namespace std;

int main()
{
int n,m,k;
while(cin>>n>>m){
if(!n&&!m) return 0;
k=pow(n,(double)1/m);
if(pow(k+1,m)-n<n-pow(k,m)) k++;//求靠近的一个
cout<<k<<endl;
}
return 0;
}