Submission #6398167

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Source Code Expand

#include <iostream>
using namespace std;
string s;//字符串 
string a[16] = {"AB","BA","AX","XA","AY","YA","BX","XB","BY","YB","XY","YX","AA","BB","XX","YY"}; //记录全部可能的组合 
int Min = 1000;
int main(){
    int n;
    cin>>n;
    cin>>s;

    string L,R;
    for(int i=0;i<16;i++){
        for(int j=0;j<16;j++){ //枚举L和R。i,j是在a字符串中的编号 
            if(i == j)//相等就跳过，因为相等绝对不会比不同好 
                continue;
            int count = 0;  
            L = a[i];
            R = a[j];

            for(int k=0;k<s.size();k++){ 
                if(s[k] == L[0]&&s[k+1] == L[1]){//如果s[k]还有s[k+1]组成的字符串等于L 
                    count++;//数量+1 
                    k++;//其实是+2但是由于for已经+1所以k++就行了 
                }
                else if(s[k] == R[0]&&s[k+1] == R[1]){//同理判断R 
                    count++;
                    k++;
                }
                else//如果不是L也不是R，那k就不加了，直接加数量 
                    count++;

            }
            if(count<Min)
                Min = count;
        }
    }
    cout<<Min<<endl;//一定要换行，不然会错 
    return 0;
}

Submission Info

Judge Result