A. Vitaly and Strings

最新推荐文章于 2019-08-29 13:21:30 发布

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本文探讨了在给定两个相同长度的字符串s和t的情况下，如何找出一个满足特定条件的字符串：该字符串的字典序大于s且小于t，并且只包含小写字母。通过分析示例输入和输出，理解并实现了解决方案。

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Vitaly is a diligent student who never missed a lesson in his five years of studying in the university. He always does his homework on time and passes his exams in time.

During the last lesson the teacher has provided two strings s and t to Vitaly. The strings have the same length, they consist of lowercase English letters, string s is lexicographically smaller than string t. Vitaly wondered if there is such string that is lexicographically larger than string s and at the same is lexicographically smaller than string t. This string should also consist of lowercase English letters and have the length equal to the lengths of strings s and t.

Let's help Vitaly solve this easy problem!

Input

The first line contains string s (1 ≤ |s| ≤ 100), consisting of lowercase English letters. Here, |s| denotes the length of the string.

The second line contains string t (|t| = |s|), consisting of lowercase English letters.

It is guaranteed that the lengths of strings s and t are the same and string s is lexicographically less than string t.

Output

If the string that meets the given requirements doesn't exist, print a single string "No such string" (without the quotes).

If such string exists, print it. If there are multiple valid strings, you may print any of them.

Sample test(s)

input
a
c

output
b

input
aaa
zzz

output
kkk

input
abcdefg
abcdefh

output
No such string做了很长时间一直WA,主要是考虑不充分。先考虑末尾的情况，然后如果前面对应第i个字符中第二串比第一串大于等于2就可以直接使得s[i]=s1[i]+1，然后其他的用'z'补齐，如果前面对应第i个字符中第二串比第一串大1就要分两种情况讨论，第一种是看s1中剩下字符是不是都为'z'，只要一个不是，s[i]=s1[i]，i后面的都用'z'补齐并输出。如果这种情况不满足，那么就看s2中是不是所有的字符都是'a'，如果有一个不是，s[i]=s2[i]，i后面的都用'a'补齐并输出。#include<stdio.h>
#include<string.h>
char s1[200],s2[200],s[200];
int main()
{
	int n,m,i,j,len,flag,flag1,flag2,flag3;
	while(scanf("%s%s",s1,s2)!=EOF)
	{
		len=strlen(s1);
		flag=1;
		flag1=0;
		memset(s,0,sizeof(s));
		for(i=0;i<len;i++){
			if(i==len-1 && s2[i]-s1[i]<=1){
				flag=0;break;
			}
			else if(i==len-1 && s2[i]-s1[i]>=2){
				s[i]=s1[i]+1;break;
			}
			else if(s2[i]-s1[i]>=2){
				s[i]=s1[i]+1;
				for(j=i+1;j<len;j++){
					s[j]='z';
				}
				break;
			}
			else if(s2[i]-s1[i]==1){
				flag2=0;
				for(j=i+1;j<len;j++){
					if(s1[j]!='z'){
						flag2=1;break;
					}
				}
				if(flag2==1){
					s[i]=s1[i];
					for(j=i+1;j<len;j++){
						s[j]='z';
					}
					break;
				}
				else if(flag2==0){
				      flag3=0; 
				      for(j=i+1;j<len;j++){
				         if(s2[j]!='a'){
      					flag3=1;break;
      				    }
				      }
      				 if(flag3==0){
 				      	flag=0;break;
 				      }
 				      else{
      				 	s[i]=s2[i];
      				 	for(j=i+1;j<len;j++){
	 				      	s[j]='a';
	 				      }
	                      break;
      				 }
				}
			}
			
			else if(s2[i]==s1[i]){
				s[i]=s1[i];continue;
			}
			else if(s1[i]>s2[i]){
				flag=0;break;
			}
			
		}
		if(flag==1)printf("%s\n",s);
		else printf("No such string\n");
	}
	return 0;
}