pta L1-064 估值一亿的AI核心代码

两要素:

1.多余空格的处理

2.读取并更改指定字符串

还得多学习

#include
#include
#include
using namespace std;
string str;
bool check(int x,int y,int len) //x为单词前一位，y为单词后一位,len为字符串长度
{
    if((x < 0|| str[x]==' '||ispunct(str[x]))&&(y>=len||str[y]==' '||ispunct(str[y])))
        return true;
    return false;
}
int main()
{
    int T;
    cin >> T;
    getchar();
    while(T--)    
    {
        getline(cin,str);
        cout << str << endl;
        int i = 0, j = str.length() - 1;
        //去掉首、尾空格
        while(str[i] == ' ') i++;
        while(str[j] == ' ') j--;
        //处理str字符串
        int cot = 0;
        for(;i <= j;i ++, cot ++) //利用双指针处理多余空格问题
        {
            if(isupper(str[i]) && str[i] != 'I')
                str[cot] = str[i] + 32;
            else if(str[i] == '?')
                str[cot] = '!';
            else if(str[i] == ' ')
            {
                str[cot] = ' ';
                while(str[++i] == ' '); //判断下一个字符是否为空格
                if(ispunct(str[i])) //符号前不需要空格
                   cot --;
                i --;
            }
            else // 不修改
                str[cot] = str[i];
        }
        string goal="";
        for(int i = 0;i < cot;i ++)
        {
            if(str[i] == 'I' && check(i - 1,i + 1, cot))
                goal += "you";
            else if(str.substr(i,2) == "me" && check(i - 1, i + 2, cot))
                goal += "you",i += 1;
            else if(str.substr(i,7) == "can you" && check(i - 1,i + 7, cot))
                goal += "I can", i += 6;
            else if(str.substr(i,9) == "could you" && check(i - 1,i + 9, cot))
                goal += "I could",i += 8;
            else goal += str[i];
        }
    goal = "AI: " + goal;
        cout << goal << endl;
    }
    return 0;
}