Project Euler：Problem 29 Distinct powers

Consider all integer combinations of a^b for 2 ≤ a ≤ 5 and 2 ≤ b ≤ 5:

2²=4, 2³=8, 2⁴=16, 2⁵=32

3²=9, 3³=27, 3⁴=81, 3⁵=243

4²=16, 4³=64, 4⁴=256, 4⁵=1024

5²=25, 5³=125, 5⁴=625, 5⁵=3125

If they are then placed in numerical order, with any repeats removed, we get the following sequence of 15 distinct terms:

4, 8, 9, 16, 25, 27, 32, 64, 81, 125, 243, 256, 625, 1024, 3125

How many distinct terms are in the sequence generated by a^b for 2 ≤ a ≤ 100 and 2 ≤ b ≤ 100?

#include <iostream>
#include <map>
#include <string>
using namespace std;

void reverse_data(string &data)
{
	char temp = '0';
	int start = 0;
	int end = data.size() - 1;

	while (start < end)
	{
		temp = data[start];
		data[start++] = data[end];
		data[end--] = temp;
	}
}
void compute_value(string lhs, string rhs, string &result)
{
	reverse_data(lhs);
	reverse_data(rhs);
	int i = 0, j = 0, res_i = 0;
	int tmp_i = 0;
	int carry = 0;

	for (i = 0; i != lhs.size(); ++i, ++tmp_i)
	{
		res_i = tmp_i;  //在每次计算时，结果存储的位需要增加
		for (j = 0; j != rhs.size(); ++j)
		{
			carry += (result[res_i] - '0') + (lhs[i] - '0') * (rhs[j] - '0');//此处注意，每次计算并不能保证以前计算结果的进位都消除， 并且以前的计算结果也需考虑。
			result[res_i++] = (carry % 10 + '0');
			carry /= 10;
		}
		while (carry)//乘数的一次计算完成，可能存在有的进位没有处理
		{
			result[res_i++] = (carry % 10 + '0');
			carry /= 10;
		}
	}
	for (int i = result.size() - 1; i >= 0; i--)
	{
		if (result[i] != '0')
			break;
		else
			result.pop_back();
	}
	reverse_data(result);
}

string powe(string a, int b)
{
	string res = a;
	for (int ii = 1; ii < b; ii++)
	{
		string tmp(res.length()+a.length(),'0');
		compute_value(res, a, tmp);
		res = tmp;
	}
	return res;
}

string int_str(int a)
{
	string s = "";
	while (a != 0)
	{
		int tp = a % 10;
		char p = tp + '0';
		s = p + s;
		a = a / 10;
	}
	return s;
}

int main()
{
	map<string, int>mp;
	for (int a = 2; a <= 100; a++)
	{
		string s = int_str(a);
		for (int b = 2; b <= 100; b++)
		{
			string res = powe(s, b);
			mp[res] = 1;
		}
	}

	map<string, int>::iterator iter;
	int count = 0;
	for (iter = mp.begin(); iter != mp.end(); iter++)
	{
		if (mp[iter->first] == 1)
			count++;
	}
	cout << count << endl;
	system("pause");
	return 0;
}