CodeForces A. Letters Cyclic Shift【字符串】

最新推荐文章于 2022-02-16 11:29:27 发布

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本文介绍了一种算法挑战，任务是在给定的字符串中选择一个非空子串，并将其中的所有字母按照英文字母表向前移动一位。目标是找到执行此操作后得到的字典序最小的字符串。

You are given a non-empty string s consisting of lowercase English letters. You have to pick exactly one non-empty substring of s and shift all its letters 'z' $\text{[math]}$ 'y' $\text{[math]}$ 'x' $\text{[math]}$ 'b' $\text{[math]}$ 'a' $\text{[math]}$ 'z'. In other words, each character is replaced with the previous character of English alphabet and 'a' is replaced with 'z'.

What is the lexicographically minimum string that can be obtained from s by performing this shift exactly once?

Input

The only line of the input contains the string s (1 ≤ |s| ≤ 100 000) consisting of lowercase English letters.

Output

Print the lexicographically minimum string that can be obtained from s by shifting letters of exactly one non-empty substring.

Examples

input
codeforces

output
bncdenqbdr

input
abacaba

output
aaacaba

Note

String s is lexicographically smaller than some other string t of the same length if there exists some 1 ≤ i ≤ |s|, such thats1 = t1, s2 = t2, ..., si - 1 = ti - 1, and si < ti.

刚做这首题的时候，宝宝也是很无语，一堆英文看不懂也就算了。。给的数据也让我分析不出来。。最后好不容易搞懂了，用JAVA写了一遍，超时，用C写，超时，最后用C++才给过了。。百度的时候发现了一个神奇的头文件#include<bits/stdc++.h>，据说包含了C++里的所有头文件，长姿势了，，，

code:

#include<bits/stdc++.h>
using namespace std;
int main() {
	string str;
	int i,j;
	cin>>str;
	for(i=0;i<str.size();i++){
		if(str[i]!='a'){
			break;
		}
	}
	for(j=i;j<str.size();j++){
		if(str[j]=='a'){
			break;
		}
		str[j]--;
	}
	if(i==str.size()){
		str[str.size()-1]='z';
	}
	cout<<str<<endl;
	return 0;
}