AtCoder Beginner Contest 082 B

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[101];
char b[101];
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[0]<b[lb-1]) cout<<"Yes"<<endl;
        else if(a[0]>b[lb-1]) cout<<"No"<<endl;
        else{
            if(a[0]==a[la-1]&&b[0]==b[lb-1]&&a[0]==b[0]){
                if(lb>la) cout<<"Yes"<<endl;
                else cout<<"No"<<endl;
                continue;
            }
            int k=0;
            for(int i=0;i<la;i++){
                for(int j=0;j<lb;j++){
                    if(b[j]>a[i]){
                        k++;
                        break;
                    }
                }
            }
            if(k==la) cout<<"Yes"<<endl;
            else cout<<"No"<<endl;
        }
    }
    return 0;
}