hello,
is anybody here who codes in C++ please? i've also got a Runtime error for the 
cryptopangan problem even if i got the same output with the sample test. that's 
my code below: 

#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;

pair<int, int> factor(int n);
int main()
{

        int T, N, L;

        vector<string> output(2);
        vector<int> input(2);
        vector<int> prime(2);

        cin >> T;
        output.resize(T);
        for (int i = 0; i < T; i++) {
                cin >> N >> L;
                input.resize(L);
                prime.resize(L + 1);
                int temp;
                for (int j = 0; j < L; j++) {
                        cin >> temp;
                        input.at(j) = temp;
                }
                pair<int, int> p;
                pair<int, int> p1;
                p = factor(input.at(0));
                p1 = factor(input.at(1));
                if ((p.first == p1.first) || (p.first == p1.second)) {
                        prime.at(0) = p.second;
                        prime.at(1) = p.first;
                        if (p.first == p1.first)
                                prime.at(2) = p1.second;
                        else prime.at(2) = p1.first;
                }
                else {
                        prime.at(0) = p.first;
                        prime.at(1) = p.second;
                        if (p1.first == p.second) prime.at(2) = p1.second;
                        else prime.at(2) = p1.first;
                }
                for (int k = 3; k < L + 1; k++) {
                        prime.at(k) = (input.at(k - 1)) / (prime.at(k - 1));
                        if (prime.at(k)>N) return 0;
                }
                vector<int> trier(L + 1);
                vector<pair<char, int>> letters(L + 1);
                pair<char, int> p2;
                trier = prime;
                sort(trier.begin(), trier.end());
                char car = 'A';
                p2 = make_pair(car, trier.at(0));
                letters.at(0) = p2;
                for (int k = 1; k < L + 1; k++) {
                        if (trier.at(k) == trier.at(k - 1))
                                letters.at(k) = letters.at(k - 1);
                        else {
                                car++;
                                p2 = make_pair(car, trier.at(k));
                                letters.at(k) = p2;
                        }
                }
                string out = "";
                for (int l = 0; l < L + 1; l++) {
                        for (int m = 0; m < L + 1; m++) {
                                if (prime.at(l) == letters.at(m).second) {
                                        out += letters.at(m).first;
                                        break;
                                }
                        }
                }
                output.at(i) = out;
        }
        for (int i = 0; i < T-1; i++) {
                cout << "Case #" << i + 1 << ": " << output.at(i) << endl;
        }
        cout << "Case #" << T << ": " << output.at(T - 1);

        //system("pause");
        return 0;
}

pair<int, int> factor(int n) {
        pair<int, int> p;
        int cpt = 0;
        for (int i = 1; i < n; ++i)
        {
                if ((n % i == 0) && (i != 1)) {
                        if (cpt == 0) {
                                p.first = i;
                                cpt++;
                        }
                        else p.second = i;
                }
        }
        return p;
}

-- 
You received this message because you are subscribed to the Google Groups 
"Google Code Jam" group.
To unsubscribe from this group and stop receiving emails from it, send an email 
to google-code+unsubscr...@googlegroups.com.
To post to this group, send email to google-code@googlegroups.com.
To view this discussion on the web visit 
https://groups.google.com/d/msgid/google-code/1df05f28-2958-4bea-b51f-bfa29690e22a%40googlegroups.com.
For more options, visit https://groups.google.com/d/optout.

Reply via email to