AtCoder Beginner Contest 082 B - Two Anagrams

题目链接：https://abc082.contest.atcoder.jp/tasks/abc082_b

Time limit : 2sec / Memory limit : 256MB

Score : 200 points

Problem Statement

You are given strings s and t, consisting of lowercase English letters. You will create a string s' by freely rearranging the characters in s. You will also create a string t' by freely rearranging the characters in t. Determine whether it is possible to satisfy s'<t' for the lexicographic order.

Notes

For a string a=a1a2…aN of length N and a string b=b1b2…bM of length M, we say a<b for the lexicographic order if either one of the following two conditions holds true:

• N<M and a1=b1, a2=b2, ..., aN=bN.
• There exists i (1≤i≤N,M) such that a1=b1, a2=b2, ..., ai−1=bi−1 and ai<bi. Here, letters are compared using alphabetical order.

For example, xy < xya and atcoder < atlas.

Constraints

• The lengths of s and t are between 1 and 100 (inclusive).
• s and t consists of lowercase English letters.

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Input

Input is given from Standard Input in the following format:

s
t

Output

If it is possible to satisfy s'<t', print Yes; if it is not, print No.

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Sample Input 1

Copy

yx
axy

Sample Output 1

Copy

Yes

We can, for example, rearrange yx into xy and axy into yxa. Then, xy < yxa.

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Sample Input 2

Copy

ratcode
atlas

Sample Output 2

Copy

Yes

We can, for example, rearrange ratcode into acdeort and atlas into tslaa. Then, acdeort < tslaa.

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Sample Input 3

Copy

cd
abc

Sample Output 3

Copy

No

No matter how we rearrange cd and abc, we cannot achieve our objective.

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Sample Input 4

Copy

w
ww

Sample Output 4

Copy

Yes

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Sample Input 5

Copy

zzz
zzz

Sample Output 5

Copy

No

补漏洞

 #include <iostream>
 #include <cstdio>
 #include <algorithm>
 #include <stack>
 #include <string>
 #include <cstring>
 #include <cmath>
 #include <cstdio>
 using namespace std;
 char a[];
 char b[];
 int main()
 {
     while(scanf("%s%s",&a,&b)!=EOF){
         int la=strlen(a);
         int lb=strlen(b);
         sort(a,a+la);
         sort(b,b+lb);
         if(a[]<b[lb-]) cout<<"Yes"<<endl;
         else if(a[]>b[lb-]) cout<<"No"<<endl;
         else{
             if(a[]==a[la-]&&b[]==b[lb-]&&a[]==b[]){
                 if(lb>la) cout<<"Yes"<<endl;
                 else cout<<"No"<<endl;
                 continue;
             }
             int k=;
             for(int i=;i<la;i++){
                 for(int j=;j<lb;j++){
                     if(b[j]>a[i]){
                         k++;
                         break;
                     }
                 }
             }
             if(k==la) cout<<"Yes"<<endl;
             else cout<<"No"<<endl;
         }
     }
     return ;
 }