题解 | #红和绿#

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

int r[52], g[52];
int min(int a, int b){return a <b?a:b;}
int main() {
    string str;
    cin >> str;
    int l = str.length();
    r[0] = g[0] = 0;
    for (int i = 1; i <= l; i++)//弄个前缀和
    {
        r[i] = r[i-1];
        g[i] = g[i-1];
        if (str[i-1] == 'R')r[i]++;
       else g[i]++; 
    }
    int ans = l;//显然最大不会超过l
    for (int i = 0; i <= l; i++)
    ans = min(ans, g[i]+r[l]-r[i]);//枚举所有位置试一下 把前面的全部刷成R，后面的刷成G
    
    cout << ans;

}