POJ-3100-Root of the Problem，原来是水题，暴力求解~~~-优快云博客

本文提供了一种通过暴力搜索解决POJ 3100问题的有效方法，该问题要求找到最接近给定数值B的N次根整数A。通过分析题目数据范围，确定了搜索区间，最终实现了一个简洁且高效的解决方案。

Root of the Problem

Time Limit: 1000MS		Memory Limit: 65536K

http://poj.org/problem?id=3100 已AK;

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

Sample Output

Source

Mid-Central USA 2006

题意应该很好看懂，给你B,K，求A^K最接近B的那个A是多少，当时想了一会，没什么思路，，差点用快速幂求了，不过认真看题发现数据范围并不大，B才在百万级内，完全可以用暴力解决，要知道2^20等于1024*1024>1000000，即最大只需for遍历到20就可以了；

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
int main()
{
    int x,k,i;
    while(~scanf("%d%d",&x,&k)&&x&&k)
    {
        if(k==1)
            printf("%d\n",x);
        else if(k>=x)//如果相等，2^k次方肯定远大于x;
            printf("1\n");
        else
        {
            for(i=1;;i++)//遍历求出满足条件A的范围；
                if((int)pow(i*1.0,k)<=x&&x<=(int)pow((i+1)*1.0,k))
                break;
            if(abs((int)pow(i*1.0,k)-x)<=abs((int)pow((i+1)*1.0,k)-x))
                printf("%d\n",i);
            else
                printf("%d\n",i+1);
        }
    }
}

转载于:https://www.cnblogs.com/nyist-TC-LYQ/p/7208267.html