我一开始还没看懂非正规数字的意义，以为那里写的单词不算，蒙了好久，而且这题非常考验仔细程度，一不小心就RE,WA。
嗯，好像讲了些废话，那我们看看思路，我的做法和前面的大佬们有些不同，因为这题只有六个字母，很多人都是单个判断它们是不是26种(one~twenty)情况中的一种，但是如果数据给的多了（超过了6个单词）感觉效率会不高；我们可以反过来。
首先，我们可以在26个单词(one~twenty)前后加一个空格，因为题目给的句子中单词前后都有空格（句子最前面要手动加一个空格。然后我们就可以用 find 函数对26种情况中的每个单词在句子中出现的次数进行累加。然后在排列输出（对应数字小的一定在前面）
下面给上代码：
#include<bits/stdc++.h>
using namespace std;
string a[27]={""," one "," two "," three "," four "," five "," six "," seven "," eight "," nine "," ten "," eleven "," twelve ", " thirteen ", " fourteen ", " fifteen ", " sixteen ", " seventeen ", " eighteen ", " nineteen ", " twenty "," a "," both "," another "," first "," second "," third "};
int s[7],i,j,b,t=1;
int main(){
getline(cin,a[0]);
a[0]=' '+a[0];//在最前面加空格。
for(i=1;i<=26;i++){
b=0;
while(a[0].find(a[i],b)!=string::npos){//搜索出现次数
t++;//出现一次，累加一次。
b=a[0].find(a[i],b)+1;//从下一个编号搜索下一个单词
}
for(;j<t;j++) s[j]=((i-1)%20+1)*((i-1)%20+1)%100;//计算对应数字
}
sort(s+1,s+t);//排序
cout<<s[1];//第一个不需要补空格
for(i=2;i<t;i++)
printf("%.2d",s[i]);//补空格输出
return 0;
}