《ACM书中题目》T

• 原题

  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.

• 理解思路
  求最逼近B的A的N次幂
  先求N次根号下B的整数部分并赋值给A，如果A的N次幂与B的差的绝对值小于A+1的N次幂与B的差的绝对值，那么A就是要求的数，否则A+1就是所要求的数。

• AC代码

#include<bits/stdc++.h>
using namespace std;
int main(){
    double b,n;
    while(cin>>b>>n,b!=0 && n!=0){
        int a= (int )pow(b,1/n);
        if(b-pow(a,n)<pow(a+1,n)-b) cout<<a<<endl;
        else cout<<a+1<<endl;
    }
    return 0;
}

• 总结
  比较简单的一个题，但是要注意数学方法的应用，否则不仅费时费力，还可能造成超时。