D. Irreducible Anagrams------------------思维(难)

Let’s call two strings s and t anagrams of each other if it is possible to rearrange symbols in the string s to get a string, equal to t.

Let’s consider two strings s and t which are anagrams of each other. We say that t is a reducible anagram of s if there exists an integer k≥2 and 2k non-empty strings s1,t1,s2,t2,…,sk,tk that satisfy the following conditions:

If we write the strings s1,s2,…,sk in order, the resulting string will be equal to s;
If we write the strings t1,t2,…,tk in order, the resulting string will be equal to t;
For all integers i between 1 and k inclusive, si and ti are anagrams of each other.
If such strings don’t exist, then t is said to be an irreducible anagram of s. Note that these notions are only defined when s and t are anagrams of each other.

For example, consider the string s= “gamegame”. Then the string t= “megamage” is a reducible anagram of s, we may choose for example s1= “game”, s2= “gam”, s3= “e” and t1= “mega”, t2= “mag”, t3= “e”:

On the other hand, we can prove that t= “memegaga” is an irreducible anagram of s.

You will be given a string s and q queries, represented by two integers 1≤l≤r≤|s| (where |s| is equal to the length of the string s). For each query, you should find if the substring of s formed by characters from the l-th to the r-th has at least one irreducible anagram.

Input
The first line contains a string s, consisting of lowercase English characters (1≤|s|≤2⋅105).

The second line contains a single integer q (1≤q≤105) — the number of queries.

Each of the following q lines contain two integers l and r (1≤l≤r≤|s|), representing a query for the substring of s formed by characters from the l-th to the r-th.

Output
For each query, print a single line containing “Yes” (without quotes) if the corresponding substring has at least one irreducible anagram, and a single line containing “No” (without quotes) otherwise.

Examples
inputCopy
aaaaa
3
1 1
2 4
5 5
outputCopy
Yes
No
Yes
inputCopy
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
outputCopy
No
Yes
Yes
Yes
No
No
Note
In the first sample, in the first and third queries, the substring is “a”, which has itself as an irreducible anagram since two or more non-empty strings cannot be put together to obtain “a”. On the other hand, in the second query, the substring is “aaa”, which has no irreducible anagrams: its only anagram is itself, and we may choose s1= “a”, s2= “aa”, t1= “a”, t2= “aa” to show that it is a reducible anagram.

In the second query of the second sample, the substring is “abb”, which has, for example, “bba” as an irreducible anagram.

题意:
求一个子串，是否存在一个同构串（字母组成相同），不可约

可约当且仅当存在一个前缀同构

解析:
输出YES一共有三种情况
第一种:长度为1
第二种:首尾不相等。例如abb,构造一个bba 是满足情况的。
第三种:字符串里面不同字符的个数>2. 例如abca 那么构造一个 caba.也是符合情况的

#include<bits/stdc++.h> 
using namespace std;
const int N=2e5+1000;
string s;
int cnt[26][N];
int t;
int l,r;
void slove()
{
	l--;r--;
	int bcnt=0;
	if(l==r)
	{
		puts("YES");
		return;
	}
	if(s[l]!=s[r])
	{
		puts("YES");
		return ;
	}
	
	for(int j=0;j<26;j++) //统计l—r有多少个 不同的字符 
	{
		int R=cnt[j][r];
		int L=0;
		if(l>0)  L=cnt[j][l-1];
		if(R-L>0) bcnt++;
	}
	if(bcnt>2) 
	{
		puts("YES");
		return ;
	}
	puts("NO");
}
int main()
{
	cin>>s;cin>>t;
	for(int i=0;i<s.size();i++) //统计前缀和。当前i位置到第0位置有多少个字符和自己一样。 
	{
		cnt[s[i]-'a'][i]++;
		if(i>0) 
		{
			for(int j=0;j<26;j++)
				cnt[j][i]+=cnt[j][i-1];
		}
	}
	while(t--)
	{
		cin>>l>>r;
		slove();
	} 
}