题解：P1603 斯诺登的密码

Andy16673328152 · 2025-07-26 07:22:58 · 题解

This is my first solution......

我的解法和别人的不同之处在于：

1. 将基数词放在一起，序数词放在一起，a 和 both 放在一起，another 单独一组，这样可以直接用下标作为转换后的数字。
2. 用 to_string(i*i%100); 来进行操作，可有效地节省空间。

Code:

#include<bits/stdc++.h>
using namespace std;
bool cmp(const string &a,const string &b){
    return (a+b<b+a);
}
string s,k,p;
string j[21]={"zero","one","two","three","four","five","six","seven","eight","nine","ten","eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen","twenty"};
string x[21]={"0th","first","second","third"};
string f[21]={"0","a","both"};
string v="another";
int a,q;
string g[7];
int main(){
    getline(cin,s);
    for(int ii=0;ii<s.size();ii++){
        if(s[ii]>='A'&&s[ii]<='Z'){
            s[ii]+=32;//注意大小写
        }
        if(s[ii]==' '){
            for(int i=0;i<=20;i++){
                if(k==j[i]||k==x[i]||k==f[i]){
                    q++;
                    g[++a]=to_string(i*i%100);
                    if(g[a].size()==1){
                        g[a].insert(0,"0");
                    }
                }
            }
            if(k==v){
                q++;
                g[++a]="01";
            }
            k="";
        }else{
            k+=s[ii];
        }
    }
    sort(g+1,g+a+1,cmp);
    for(int i=1;i<=a;i++){
        p+=g[i];
    }
    while(p[0]=='0'){
        p.erase(0,1);
    }
    if(q==0){//如果没有合法的数字
        cout<<q;
    }else{
        cout<<p;
    }
    return 0;
}