模拟题。
先写一个函数 \(\textrm{win}(a,b)\) 表示两个糖果 \(a\) 和 \(b\) 哪个会赢，方便后续判断操作。然后记录两个指针 \(p,q\) 表示 Ngoc 和 Minh 队伍当前最前面的糖果是哪一个，每次比较最前面的糖果即可。
时间复杂度 \(O(N+M)\)。
代码：
// Problem: P15538 [CCC 2026 J3] Creative Candy Consumption
// Contest: Luogu
// URL: https://www.luogu.com.cn/problem/P15538
// Memory Limit: 512 MB
// Time Limit: 1000 ms
//
// Powered by CP Editor (https://cpeditor.org)//By: OIer rui_er
#include <bits/stdc++.h>
#define rep(x, y, z) for(int x = (y); x <= (z); ++x)
#define per(x, y, z) for(int x = (y); x >= (z); --x)
#define debug(format...) fprintf(stderr, format)
#define fileIO(s) do {freopen(s".in", "r", stdin); freopen(s".out", "w", stdout);} while(false)
#define endl '\n'
using namespace std;
typedef long long ll;mt19937 rnd(std::chrono::duration_cast<std::chrono::nanoseconds>(std::chrono::system_clock::now().time_since_epoch()).count());
int randint(int L, int R) {uniform_int_distribution<int> dist(L, R);return dist(rnd);
}template<typename T> void chkmin(T& x, T y) {if(y < x) x = y;}
template<typename T> void chkmax(T& x, T y) {if(x < y) x = y;}template<int mod>
inline unsigned int down(unsigned int x) {return x >= mod ? x - mod : x;
}template<int mod>
struct Modint {unsigned int x;Modint() = default;Modint(unsigned int x) : x(x) {}friend istream& operator>>(istream& in, Modint& a) {return in >> a.x;}friend ostream& operator<<(ostream& out, Modint a) {return out << a.x;}friend Modint operator+(Modint a, Modint b) {return down<mod>(a.x + b.x);}friend Modint operator-(Modint a, Modint b) {return down<mod>(a.x - b.x + mod);}friend Modint operator*(Modint a, Modint b) {return 1ULL * a.x * b.x % mod;}friend Modint operator/(Modint a, Modint b) {return a * ~b;}friend Modint operator^(Modint a, int b) {Modint ans = 1; for(; b; b >>= 1, a *= a) if(b & 1) ans *= a; return ans;}friend Modint operator~(Modint a) {return a ^ (mod - 2);}friend Modint operator-(Modint a) {return down<mod>(mod - a.x);}friend Modint& operator+=(Modint& a, Modint b) {return a = a + b;}friend Modint& operator-=(Modint& a, Modint b) {return a = a - b;}friend Modint& operator*=(Modint& a, Modint b) {return a = a * b;}friend Modint& operator/=(Modint& a, Modint b) {return a = a / b;}friend Modint& operator^=(Modint& a, int b) {return a = a ^ b;}friend Modint& operator++(Modint& a) {return a += 1;}friend Modint operator++(Modint& a, int) {Modint x = a; a += 1; return x;}friend Modint& operator--(Modint& a) {return a -= 1;}friend Modint operator--(Modint& a, int) {Modint x = a; a -= 1; return x;}friend bool operator==(Modint a, Modint b) {return a.x == b.x;}friend bool operator!=(Modint a, Modint b) {return !(a == b);}
};int win(char a, char b) {if(a == b) return 0;if(a == 'R' && b == 'G') return +1;if(a == 'G' && b == 'B') return +1;if(a == 'B' && b == 'R') return +1;return -1;
}int main() {ios::sync_with_stdio(false);cin.tie(0); cout.tie(0);string s, t;cin >> s >> t;int n = s.size(), m = t.size();int p = 0, q = 0, ansp = 0, ansq = 0;while(p < n && q < m) {if(win(s[p], t[q]) == 0) {++p;++q;++ansp;++ansq;}else if(win(s[p], t[q]) == +1) {++q;++ansp;}else {++p;++ansq;}}ansp += n - p;ansq += m - q;cout << ansp << endl << ansq << endl;return 0;
}