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 stringssandtto Vitaly. The strings have the same length, they consist of lowercase English letters, stringsis lexicographically smaller than stringt. Vitaly wondered if there is such string that is lexicographically larger than stringsand at the same is lexicographically smaller than stringt. This string should also consist of lowercase English letters and have the length equal to the lengths of stringssandt.
Let’s help Vitaly solve this easy problem!
Input
The first line contains strings(1≤|s|≤100), consisting of lowercase English letters. Here,|s|denotes the length of the string.
The second line contains stringt(|t|=|s|), consisting of lowercase English letters.
It is guaranteed that the lengths of stringssandtare the same and stringsis lexicographically less than stringt.
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
Strings=s1s2…snis said to be lexicographically smaller thant=t1t2…tn, if there exists suchi, thats1=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;}
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