cf397 B. Code obfuscation

B. Code obfuscation
time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
Kostya likes Codeforces contests very much. However, he is very disappointed that his solutions are frequently hacked. That’s why he decided to obfuscate (intentionally make less readable) his code before upcoming contest.

To obfuscate the code, Kostya first looks at the first variable name used in his program and replaces all its occurrences with a single symbol a, then he looks at the second variable name that has not been replaced yet, and replaces all its occurrences with b, and so on. Kostya is well-mannered, so he doesn’t use any one-letter names before obfuscation. Moreover, there are at most 26 unique identifiers in his programs.

You are given a list of identifiers of some program with removed spaces and line breaks. Check if this program can be a result of Kostya’s obfuscation.

Input
In the only line of input there is a string S of lowercase English letters (1 ≤ |S| ≤ 500) — the identifiers of a program with removed whitespace characters.

Output
If this program can be a result of Kostya’s obfuscation, print “YES” (without quotes), otherwise print “NO”.

Examples
input
abacaba
output
YES
input
jinotega
output
NO
Note
In the first sample case, one possible list of identifiers would be “number string number character number string number”. Here how Kostya would obfuscate the program:

replace all occurences of number with a, the result would be “a string a character a string a”,
replace all occurences of string with b, the result would be “a b a character a b a”,
replace all occurences of character with c, the result would be “a b a c a b a”,
all identifiers have been replaced, thus the obfuscation is finished.
题意：其实就是按a,b,c,d……….的大小排列,必须按顺序，能出现比它上一个最大字母小的，但不能比它大。

#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<iostream>
#include<algorithm>
using namespace std;
char str[505];
int main()
{
    scanf("%s",str);
    int len=strlen(str);
    char a='a';//赋值时，字母要加单引号
    int flag=0;
    for(int i=0;i<len;i++)
    {
        if(str[i]==a)//这两个地方不用加单引号，否则表示字母a，也就是数字97
        {
            a=a+1;
        }
        else if(str[i]<a)
        {
            continue;
        }
        else if(str[i]>a)
        {
            flag=1;
            printf("NO
");
            break;
        }
    }
    if(flag==0)
    {
        printf("YES
");
    }
    return 0;
}