D. Irreducible Anagrams
time limit per test2 seconds
memory limit per test256 megabytes
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
input
aaaaa
3
1 1
2 4
5 5
output
Yes
No
Yes
input
aabbbbbbc
6
1 2
2 4
2 2
1 9
5 7
3 5
output
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.
定义anagram，表示两个字符串的字符组成相同。
定义reducible anagram，表示可以两个字符串可以拆分成k个子串，且每个子串都是anagram的。
irreducible anagram 就是不满足条件的。
现在给你一个字符串，然后q次询问，每次询问给你l，r。问你[l,r]的这个字符串，能否找到一个irreducible anagram。
只有三种情况能够找到。
#include<bits stdc++.h>
using namespace std;
string s;
int q;
int cnt[26][200005];
int main(){
cin>>s;
cin>>q;
for(int i=0;i<s.size();i++){ cnt[s[i]-'a'][i]++; if(i>0){
for(int j=0;j<26;j++){ cnt[j][i]+="cnt[j][i-1];" } for(int i="0;i<q;i++){" int l,r; cin>>l>>r;l--,r--;
if(l==r){
cout<<"yes"<<endl; continue; } if(s[l]!="s[r]){" cout<<"yes"<<endl; int cnt="0;" for(int j="0;j<26;j++){" r="cnt[j][r];" l="0;" if(l>0)L=cnt[j][l-1];
if(R-L>0)Cnt++;
}
if(Cnt>2){
cout<<"yes"<<endl; continue; } cout<<"no"<<endl; < code></"yes"<<endl;></"yes"<<endl;></26;j++){></s.size();i++){></bits>
Original: https://www.cnblogs.com/qscqesze/p/12256902.html
Author: qscqesze
Title: Codeforces Round #616 (Div. 2) D. Irreducible Anagrams 找规律
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