Codeforces Round #293 (Div. 2)A. Vitaly and Strings

最新推荐文章于 2022-11-22 23:15:14 发布

Anjone_

最新推荐文章于 2022-11-22 23:15:14 发布

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分类专栏： Codeforces题解报告文章标签： codeforces

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本文探讨了一道有趣的字符串问题，即寻找一个字典序介于两个给定字符串之间的字符串。通过逐步递减扫描和字符替换的方法，给出了简洁有效的解决策略。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

这题非常有意思，各种HACk - - ，给HACK后正场比赛都。。。。。还花了大把的时间去改，还WA ，水平太差

题目连接：http://codeforces.com/contest/518/problem/A

题意：给出两个字符串s、t （t字典序比s大），问有没有字典序在s 和t 之间的字符串有就输出任意一个符合条件的，没有输出 "No such string"

想想挺简单的，要是我们用手写肯定100%写出正确答案，可是码代码，我就弱了。

思路：从s串的末端开始往前扫，我们的目的是按字典序找出在s串后一个的字符串，然后与t串比较看看是否相等，相等就GG，不相等就输出s,这里string比较可以直接用 == 判断写起来挺方便的。

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

#include<iostream>
using namespace std;
int main()
{
     string s,t;
     cin >> s >> t;
     for(int i = s.length()-1 ; i >= 0 ; i--)
     {
          if(s[i] != 'z')
          {
               s[i]++;
               break;
          }
          else
               s[i] = 'a';
     }
     if(s!=t)
          cout << s;
     else
          cout << "No such string" << endl;
     return 0;
}