#include <iostream>
#include <cmath>
using namespace std;
void main(){
    int a[100001];
    char s[100001];
    gets_s(s);
    int max = 0;
    for (int i = 0; i <= strlen(s); i++)
    {
        if (s[i] >= 'A'&& s[i] <= 'Z'){
            a[i] = s[i] - 55;
            if (max < a[i]){
                max = a[i];
            }
        }
        else{////if digits
            a[i] = s[i] - 48;
            if (max < a[i]){
                max = a[i];
            }
        };
    };
    int k;
    long long res;
    max++;
    bool found = 1;
    for (k = max; k <= 36; k++)
    {
        res = 0;
        for (int i = strlen(s) - 1; i >= 0; i--)
        {
            res = res + pow(k, i)*a[i];
        }
        if (res % (k - 1) == 0){
            found = 0;
            break;
        }
    }
    if (found==0){
        cout << k;
    }
    else{
        cout << "No solution.";
    }
    system("pause");
}
I think that problem in large numbers in tests, am I right?
Edited by author 09.07.2016 22:05

Edited by author 09.07.2016 22:26

According to the description the input can have 10^6 digits this means that the input string can be a million characters long. This will definitely overflow even "long long" :)

There is a solution to this problem described below. They shown a proof that all the calculations that You need to do is summing up the digits. I'm experimenting with a different approach. I try to simulate "pen and paper" division, and get the remainder.

And finally I got AC with my approach.

Edited by author 12.07.2016 14:48