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Black Obama is two five zero .

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You are a three eight pig .

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if(num[0]) cout<<num[0];
for(int i=1;i<cnt;i++){
    if(num[i]>10) cout<<num[i];
    else cout<<'0'<<num[i];
}

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if(!cnt) cout<<0,exit(0);

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#include<bits/stdc++.h>
using namespace std;
string tmp,dic[27]={"","one","two","three","four","five",//并不会出现zero 
"six","seven","eight","nine","ten",//但是这里我为了前20个数下标对应数值把0空着
"eleven","twelve","thirteen","fourteen","fifteen",
"sixteen","seventeen","eighteen","nineteen","twenty",
"a","both","another","first","second","third"};
int num[6],cnt;
int main(){
    for(int i=0;i<6;i++){
        cin>>tmp;
        if(!i&&tmp[0]>='A'&&tmp[0]<='Z') tmp[0]+='a'-'A';
        for(int j=1;j<=26;j++){
            if(tmp==dic[j]){
                if(j<=20){			//在下标1-20匹配即对应1-20 
                    num[cnt++]=j;
                    break;
                } else {
                    switch(j){
                    case 21: case 23: case 24:	//21、23、24对应1 
                        num[cnt++]=1;break;
                    case 22: case 25:		//22、25对应2 
                        num[cnt++]=2;break;
                    case 26:			//26对应3
                        num[cnt++]=3;break;
                    }
                    break;//switch 语句内的 break 仅能跳出 switch 而对外面的循环没有影响，
                    //因此必须在跳出 switch 后再 break 一次以确保跳出循环（其实继续遍历应该
                    //也没事，不会成功匹配只是浪费点时间）
                }
            }
        }
    }
    if(!cnt) cout<<0,exit(0);//冷知识，虽然 return 0 不能写在逗号后面用于结束程序但是
			     // exit(0) 可以
    for(int i=0;i<cnt;i++){
        num[i]=num[i]*num[i];
        num[i]%=100;
    }
    sort(num,num+cnt);
    if(num[0]) cout<<num[0];
    for(int i=1;i<cnt;i++){
        if(num[i]>10){
            cout<<num[i];
        } else {
            cout<<'0'<<num[i];
        }
    }
}