poj_3100_Root_of_the_Problem.cpp

Problem Links:

poj3100,

Problem:

Root of the Problem

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 8575		Accepted: 4589

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

4 3
5 3
27 3
750 5
1000 5
2000 5
3000 5
1000000 5
0 0

Sample Output

1
2
3
4
4
4
5
16

Source

Mid-Central USA 2006

Solution:

Easy implementation. But take care.

Source Code:

[sourcecode language="cpp" collapse="true" padlinenumbers="true"]
//Sun 31 Jan 2010 04:43:48 PM CST
#include <iostream>
#include <string>
#include <vector>
#include <cmath>

using namespace std;

int main()
{
int B, N;
cin >> B >> N;
while(B + N != 0)
{
int ret = (int) pow(B, 1.0/N);
int res1 = (int) pow(ret, 1.0*N);
int res2 = (int) pow(ret+1, 1.0*N);
if(res1+res2 > 2*B) cout << ret << endl;
else cout << ret+1 << endl;
cin >> B >> N;
}
return 0;
}
[/sourcecode]