B. Cards

// Problem: C. K-special Tables
// Contest: Codeforces - Codeforces Round #342 (Div. 2)
// URL: https://codeforces.com/problemset/problem/625/C
// Memory Limit: 256 MB
// Time Limit: 2000 ms
// 2022-03-18 15:13:59
// 
// Powered by CP Editor (https://cpeditor.org)

#include<bits/stdc++.h>
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef vector<int> vi;   
typedef vector<ll> Vll;               
typedef vector<pair<int, int> > vpii;
typedef vector<pair<ll, ll> > vpll;                        

const ll mod = 1e9 + 7;
//const ll mod = 998244353;
const double pi  = acos(-1.0);

inline ll qmi (ll a, ll b) {
    ll ans = 1;
    while (b) {
        if (b & 1) ans = ans * a%mod;
        a = a * a %mod;
        b >>= 1;
    }
    return ans;
}
inline int read () {
    int x = 0, f = 0;
    char ch = getchar();
    while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
    while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
    return f?-x:x;
}
template<typename T> void print(T x) {
    if (x < 0) putchar('-'), x = -x;
    if (x >= 10) print(x/10);
    putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
    return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
    return (a + b) %mod;
}
inline ll inv (ll a) {
    return qmi(a, mod - 2);
}
string ans;

void solve() {
    int n, k;
    cin >> n;
    string s;
    cin >> s;
    int kind = 0;
    map<char, int> cnt;
    
    for (auto t : s) {
        cnt[t] ++;
        if (cnt[t] == 1) {
            kind ++;
        }
    }
    
    if (kind == 3) {
        cout << "BGR" << endl;
    }
    else if (kind == 2) {
        if (n == 2) {
            if (cnt['R'] == 0) {
                cout <<  "R" << endl;
            }
            else if (cnt['B'] == 0) {
                cout << "B" << endl;
            }
            else cout << "G" << endl;
            return;
        }
        string ans = "";
        if (cnt['R'] == 0)
        {
            if (cnt['G'] >= 2 && cnt['B'] >= 2) {
                ans += "BGR";
            }
            else {
            ans += "R";
            if (cnt['G'] > cnt['B'])
                ans += "B";
            else 
                ans += "G";
            }
        }
        else if (cnt['B'] == 0)
        {
            if (cnt['G'] >= 2 && cnt['R'] >= 2) {
                ans += "BGR";
            }
            else {
            ans += "B";
            
            if (cnt['G'] > cnt['R'])
                ans += "R";
            else ans += "G";}
        }
        else {
            if (cnt['R'] >= 2 && cnt['B'] >= 2) {
                ans += "BGR";
            }
            else {
                ans += "G";
                if (cnt['B'] > cnt['R'])
                    ans += "R";
                else 
                    ans += "B";
            }
        }
        sort(all(ans));
        cout << ans << endl;
        
    }
    else {
        cout <<s[0] << endl;
    }
}
int main () {
    // ios::sync_with_stdio(0),cin.tie(0), cout.tie(0);
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

题意:给定变换规则，求最后剩下的有几种情况

从数量的角度去找规律，能够统一所有情况，我一开始是在具体的样例找规律