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

A. 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

ac

output

b

input

aaazzz

output

kkk

input

abcdefgabcdefh

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：题目类似于大数的进位与比较，只要将第一个字符串'+1'再与第二个串比较即可，注意最好先加1再比较，如果先比较则更麻烦，，因为要考虑aaazzz和aabaaa之类的数据

AC代码：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;char s[105], t[105];int main() {while(scanf("%s %s", s, t) != EOF) {int up, i, len = strlen(s);for(i = 0; i < len; i++) {if(s[i] != t[i]) break;}s[len - 1]++;if(s[len - 1] - 'z' == 1) {up = 1; s[len - 1] = 'a'; }else up = 0;for(int j = len - 2; j >= 0; j--) {s[j]+=up;if(s[j] - 'z' == 1) {up = 1; s[j] = 'a'; }else up = 0;}if(strcmp(s, t) == 0) printf("No such string\n"); else printf("%s\n", s);}return 0;}