Atcoder：TrBBnsformBBtion（思维 & 字符串）

E - TrBBnsformBBtion

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Time limit : 2sec / Memory limit : 256MB

Score : 600 points

Problem Statement

Let us consider the following operations on a string consisting of A and B:

1. Select a character in a string. If it is A, replace it with BB. If it is B, replace with AA.
2. Select a substring that is equal to either AAA or BBB, and delete it from the string.

For example, if the first operation is performed on ABA and the first character is selected, the string becomes BBBA. If the second operation is performed on BBBAAAA and the fourth through sixth characters are selected, the string becomes BBBA.

These operations can be performed any number of times, in any order.

You are given two string S and T, and q queries ai,bi,ci,di. For each query, determine whether SaiSai+1…Sbi, a substring of S, can be made into TciTci+1…Tdi, a substring of T.

Constraints

• 1≤|S|,|T|≤105
• S and T consist of letters A and B.
• 1≤q≤105
• 1≤ai≤bi≤|S|
• 1≤ci≤di≤|T|

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Input

Input is given from Standard Input in the following format:

STqa1 b1 c1 d1…aq bq cq dq

Output

Print q lines. The i-th line should contain the response to the i-th query. If SaiSai+1…Sbi can be made into TciTci+1…Tdi, print YES. Otherwise, print NO.

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Sample Input 1

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BBBAAAABABBBBA47 9 2 57 9 1 41 7 2 51 7 2 4

Sample Output 1

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YESNOYESNO

The first query asks whether the string ABA can be made into BBBA. As explained in the problem statement, it can be done by the first operation.

The second query asks whether ABA can be made into BBBB, and the fourth query asks whether BBBAAAA can be made into BBB. Neither is possible.

The third query asks whether the string BBBAAAA can be made into BBBA. As explained in the problem statement, it can be done by the second operation.

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Sample Input 2

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AAAAABBBBAAABBBBAAAABBBBAAABBBBBBAAAAABB102 15 2 132 13 6 161 13 2 204 20 3 201 18 9 192 14 1 113 20 3 156 16 1 174 18 8 207 20 3 14

Sample Output 2

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YESYESYESYESYESYESNONONONO

题意：给定两个字符串s和t，有两个变换规则，A变成BB或B变成AA，删除AAA或删除BBB，给出q个询问，每个询问给a,b,c,d判断s串中a到b的子串能否通过变换成t串种c到d对的子串。

思路：通过写写画画发现A可以变成AAAA，但A，AA，AAA却不能互变，于是将字符串变成全A串，模3看看是否一样就行了。

# include <bits/stdc++.h>using namespace std;const int N = 1e5+3;char s[N], t[N];int sp[N], tp[N];int main(){    int q, a, b, c,d;    scanf("%s%s%d",s+1,t+1,&q);    for(int i=1; s[i]; ++i)        sp[i] = sp[i-1]+s[i]-'A'+1;    for(int i=1; t[i]; ++i)        tp[i] = tp[i-1]+t[i]-'A'+1;    while(q--)    {        scanf("%d%d%d%d",&a,&b,&c,&d);        if((sp[b]-sp[a-1])%3 == (tp[d]-tp[c-1])%3)            puts("YES");        else            puts("NO");    }    return 0;}