学步园

//其实是一道很水的题，点知道我在没有读懂题意的情况之下，把其做法往深处想了！
//注意：You need to expand at most one prefix and one suffix，意思是说只要进行一次前缀和后缀的
//转换即可（要在有前缀和后缀的情况之下啊） 
#include <iostream>
#include <string>
#include <map> 
using namespace std;

string pre[5], suf[5];
map<string, string> m; 

int main()
{
    int i, j, n, pos, prenum;
    bool flag; 
    string word, tmp, preans, sufans, ans; 
    pre[0] = "anti", pre[1] = "post", pre[2] = "pre", pre[3] = "re", pre[4] = "un";
    suf[0] = "er", suf[1] = "ing", suf[2] = "ize", suf[3] = "s", suf[4] = "tion";
    m[pre[0]] = "against " , m[pre[1]] = "after ", m[pre[2]] = "before ", 
    m[pre[3]] = " again", m[pre[4]] = "not ", m[suf[0]] = "one who ", m[suf[1]] = "to actively ", 
    m[suf[2]] = "change into ", m[suf[3]] = "multiple instances of ", m[suf[4]] = "the process of "; 
    cin >> n; 
    while (n--){
          cin >> word; 
          preans.clear();
          sufans.clear();
          ans.clear(); 
          flag = false;  
          for (i = 0; i < 5; i++){
                pos = word.find(pre[i]); 
                if (pos == 0){
                    if (i != 3)
                        preans += m[pre[i]];
                    else{//如果前缀是re的，要记录出现过，到最后再作处理 
                         flag = true;
                    } 
                    word.erase(0, pre[i].length());
                    break; 
                }
          }
          for (i = 0; i < 5; i++){ 
                pos = word.rfind(suf[i]);
                if (pos >= 0 && pos == word.length()-suf[i].length()){
                    sufans += m[suf[i]];
                    word.erase(pos, suf[i].length());
                    if (i == 0)
                        word += "s";
                    else if (i == 4)
                        word += "ing";
                    break; 
                }
          }
          sufans += word; 
          ans = preans + sufans; 
          if (flag){
               ans += " again"; 
          } 
          cout << ans << endl; 
           
    } 
    
    system("pause"); 
}