DMOJ · ccc26j3

CCC 2026 J3 Creative Candy Consumption

This C++ solution uses string processing for CCC 2026 J3 Creative Candy Consumption. Read the reasoning, inspect the code, or try your own test case below.

ccc26j3CCC 2026 J3MathString processingC++48 lines
Solution067of 248
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Approach

String processing

Creative Candy Consumption gives two RGB queues and rock-paper-scissors consumption rules; the code simulates them and outputs each person's total.

Math

Problem and code

Useful links.

Written by benbenyaojifen. Try the problem first, then compare your approach with the code.

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Implementation

J_3_Creative_Candy_Consumption.cpp

C++

    #include <bits/stdc++.h>
    using namespace std;
    using ll = long long;
    using i128 = __int128;
    const int inf = 1e9;
    const ll INF = 1LL << 60; //❄️
    int main() {
        ios::sync_with_stdio(0); cin.tie(0); 
        string a, b; 
        getline(cin, a);
        getline(cin, b);
        int i = 0, j = 0, n = 0, m = 0;
        while (1) {
            if (i == a.size()) {
                m += b.size() - j;
                break;
            }
            if (j == b.size()) {
                n += a.size() - i;
                break;
            }
            if (a[i] == b[j]) {
                n++; m++;
                i++; j++;
            } else if (a[i] == 'R' && b[j] == 'G') {
                n++;
                j++;
            } else if (a[i] == 'G' && b[j] == 'R') {
                m++;
                i++;
            } else if (a[i] == 'R' && b[j] == 'B') {
                m++;
                i++;
            } else if (a[i] == 'B' && b[j] == 'R') {
                n++;
                j++;
            } else if (a[i] == 'B' && b[j] == 'G') {
                m++;
                i++;
            } else if (a[i] == 'G' && b[j] == 'B') {
                n++;
                j++;
            }
        }
        cout << n << '\n';
        cout << m << '\n';
        return 0;
    }
        

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