zoj 2818 Root of the Problem

标签：find   ace   math   pairs   ret   mil   title   while   pair   

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.

 1 #include <iostream>
 2 #include <cmath>
 3 #include <cstdlib>
 4 using namespace std;
 5 int main(){
 6     double b, n;
 7     while(cin >> b >> n){
 8         if(b == 0 && n == 0)
 9             break;
10         //pow:double pow(double, double)
11         int a = (int)pow(b, 1/ n);//直接取整，不是四舍五入 ,pow(a, n) < b
12         if(b - pow(a, n) < pow(a + 1, n) - b)
13             cout << a << endl;
14         else
15             cout << a + 1 << endl;
16     }
17     return 0;
18 }

zoj 2818 Root of the Problem

标签：find   ace   math   pairs   ret   mil   title   while   pair   

原文地址：http://www.cnblogs.com/qinduanyinghua/p/6529936.html