CodeForces 518A

http://acm.hust.edu.cn/vjudge/contest/view.action?cid=120930#problem/A

A - Chewbaсca and Number

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 Input

Input

a
c

Output

b

Input

aaa
zzz

Output

kkk

Input

abcdefg
abcdefh

Output

No such string

Hint

String s = s₁s₂... s_n is said to be lexicographically smaller than t = t₁t₂... t_n, if there exists such i, that s₁ = t₁, s₂ = t₂, ... s_i - 1 = t_i - 1, s_i < t_i.

题意:给出两个长度相同的字符串,问是否存在字典序在两个串之间的,存在则出,不存在输出 No such string;

wrong 原因:思考不够全面,某些数据(@1)不能通过

思路1:(较繁琐)排头遍历找到某个位置两个字母相差>1,则s[i]=s[i]+1,输出s,

若相差等于1,则把s[i+1]以及s[i+1]之后的位置都变成'z',

如果改变之后的s跟以前的s相同,则令s[i]=t[i]把s[i+1]以及s[i+1]之后的位置都变成'a',只要s!=t则输出s否则输出No such string

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
using namespace std;
typedef long long ll;

int main(){
    string a,b,c;
    while(cin>>a){
        cin>>b;
        c=a;
        int len=a.size();
        int flag=1;
        for(int i=0;i<len;i++){
            if(a[i]<b[i]){
                if(b[i]-a[i]>1){
                    a[i]=a[i]+1;
                    flag=0;
                    break;
                }
                else {
                    if(i==len-1)break;
                    else{
                        for(int j=i+1;j<len;j++){
                            a[j]='z';
                        }
                        if(a==c){
                            a[i]=b[i];
                            for(int j=i+1;j<len;j++){
                                a[j]='a';
                            }
                        }
                        if(a==b)break;
                        else {flag=0;break;}
                    }
                }
            }
        }
        if(!flag)cout<<a<<endl;
        if(flag)cout<<"No such string\n";
    }

    return 0;
}

思路2:看成26进制的数,把s+1如果得到的s在原来的s和t之间则输出注意进位问题

#include<iostream>
#include<iomanip>
#include<cstdio>
#include<string>
using namespace std;
typedef long long ll;

int main(){
    string a,b,c;
    while(cin>>a){
        cin>>b;
        c=a;
        for(int i=a.size()-1;i>=0;--i){
            if(a[i]=='z')a[i]='a';
            else {a[i]=a[i]+1; break;}
        }
        if(a!=c&&a!=b)cout<<a<<endl;
        else cout<<"No such string\n";
    }
    return 0;
}