Disarium Number (Special Numbers Series #3)

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Definition

Disarium number is the number that The sum of its digits powered with their respective positions is equal to the number itself.

Task

Given a number, Find if it is Disarium or not .

#include <string>
#include <cmath>

using namespace std; 

string int2str(int n)
{
    int m = n;
    char s[100];
    char ss[100];
    int i = 0, j = 0;
    if (n < 0)
    {
        m = 0 - m;
        j = 1;
        ss[0] = '-';
    }
    while (m > 0)
    {
        s[i++] = m % 10 + '0';
        m /= 10;
    }
    s[i] = '\0';
    i = i - 1;
    while (i >= 0)
    {
        ss[j++] = s[i--];
    }
    ss[j] = '\0';
    string res = ss;
    return res;
}

int char2int(char ch)
{
    switch (ch)
    {
        case '0': return 0; break;
        case '1': return 1; break;
        case '2': return 2; break;
        case '3': return 3; break;
        case '4': return 4; break;
        case '5': return 5; break;
        case '6': return 6; break;
        case '7': return 7; break;
        case '8': return 8; break;
        case '9': return 9; break;
    }
}

string disariumNumber (int number )
{
    string res = int2str(number);
    int n = 0;
    for (int i = 1; i <= res.size(); i++) {
        n += pow(char2int(res[i - 1]), i);
    }
    if (n == number) return "Disarium !!";
    return "Not !!";
}