CF Vitaly and Strings

Vitaly and Strings
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

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
Note

String s = s1s2... sn is said to be lexicographically smaller than t = t1t2... tn, if there exists such i, that s1 = t1, s2 = t2, ... si - 1 = ti - 1, si < ti.

 

 

 

 

思路:将字符串看成数字,先从左向右找到第一个不相同的,然后把最右边那位加一,如果超出了Z就进位。

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 #include <string>
 5 #include <algorithm>
 6 #include <cctype>
 7 #include <queue>
 8 #include <map>
 9 using    namespace    std;
10 
11 int    main(void)
12 {
13     string    s;
14     string    t;
15     int    loc,len,flag;
16 
17     cin >> s >> t;
18     if(s == t)
19         puts("No such string");
20     else
21     {
22         len = s.size();
23         flag = 0;
24 
25         for(int i = 0;i < len;i ++)
26             if(s[i] < t[i])
27             {
28                 loc = i;
29                 break;
30             }
31         if(s[loc] + 1 < t[loc])
32         {
33             s[loc] += 1;
34             flag = 1;
35         }
36         else
37         {
38             int    box = 1;
39             int    add = 0;
40             for(int i = len - 1;i > loc;i --)
41             {
42                 flag = 1;
43                 s[i] += box + add;
44                 if(s[i] > 'z')
45                 {
46                     s[i] = 'a';
47                     add = 1;
48                     box = 0;
49                 }
50                 else
51                 {
52                     add = 0;
53                     box = 0;
54                 }
55             }
56             s[loc] += add;
57             if(s == t)
58             {
59                 puts("No such string");
60                 return    0;
61             }
62         }
63         if(flag)
64             cout << s << endl;
65         else
66             puts("No such string");
67     }
68 
69     return    0;
70 }

 

posted @ 2015-04-12 23:02  Decouple  阅读(314)  评论(0编辑  收藏  举报