题目：http://poj.org/problem?id=2109

Description

Current work in cryptography involves (among other things) large prime numbers and computing powers of numbers among these primes. Work in this area has resulted in the practical use of results from number theory and other branches of mathematics once considered to be only of theoretical interest. 

This problem involves the efficient computation of integer roots of numbers. 

Given an integer n>=1 and an integer p>= 1 you have to write a program that determines the n th positive root of p. In this problem, given such integers n and p, p will always be of the form k to the n
^th. power, for an integer k (this integer is what your program must find).

Input

The input consists of a sequence of integer pairs n and p with each integer on a line by itself. For all such pairs 1<=n<= 200, 1<=p<10
¹⁰¹ and there exists an integer k, 1<=k<=10
⁹ such that k
ⁿ = p.

Output

For each integer pair n and p the value k should be printed, i.e., the number k such that k n =p.

Sample Input

2 16
3 27
7 4357186184021382204544

Sample Output

4
3
1234

基本思路：

根据p的位数基本确定k的位数，例如上述例子的最后一个数字4357186184021382204544共22位，除以7取正为3，所以k的位数应该为3。之后我首先考虑的是两分法，但是结果是超时；于是我采用的是逐步确定每一位的数字并且避免做除法运算。

代码：

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

int GetWeight(double num)
{
int result = 0;
while((num/=10) > 1)
{
result++;
};
return result + 1;
}

double GetPow(double value, int weight)
{
double result = 1;
double cur = value;
do
{
if(weight % 2 == 1)
{
result *= cur;
}
cur*=cur;
}
while((weight=weight/=2)!=0);
return result;
}

int main()
{
int n;
double p;
while(cin>>n>>p)
{
if(p == 1)
{
cout<<1;
}
else
{
int w = GetWeight(p);
int weight = w/n;
int left = w – weight * n;

double value = 0;
for(int j = weight; j >= 0; j–)
{
double tmp = GetPow(10,j);
int i =1;
for(; i <= 9; i++)
{
if(GetPow(value + i * tmp, n)>p)
{
break;
}
else if(GetPow(value + i * tmp, n)==p)
{
i++;
break;
}
}
value += (i-1) * tmp;
}
cout<<floor(value)<<endl;
}
};

return 0;
}