【Codeforces #470 Div. 2 E】Picking Strings

problem

outputstandard output
Alice has a string consisting of characters ‘A’, ‘B’ and ‘C’. Bob can use the following transitions on any substring of our string in any order any number of times:

A BC
B AC
C AB
AAA empty string
Note that a substring is one or more consecutive characters. For given queries, determine whether it is possible to obtain the target string from source.

Input
The first line contains a string S (1 ≤ |S| ≤ 105). The second line contains a string T (1 ≤ |T| ≤ 105), each of these strings consists only of uppercase English letters ‘A’, ‘B’ and ‘C’.

The third line contains the number of queries Q (1 ≤ Q ≤ 105).

The following Q lines describe queries. The i-th of these lines contains four space separated integers ai, bi, ci, di. These represent the i-th query: is it possible to create T[ci..di] from S[ai..bi] by applying the above transitions finite amount of times?

Here, U[x..y] is a substring of U that begins at index x (indexed from 1) and ends at index y. In particular, U[1..|U|] is the whole string U.

It is guaranteed that 1 ≤ a ≤ b ≤ |S| and 1 ≤ c ≤ d ≤ |T|.

Output
Print a string of Q characters, where the i-th character is ‘1’ if the answer to the i-th query is positive, and ‘0’ otherwise.

Example
inputCopy
AABCCBAAB
ABCB
5
1 3 1 2
2 2 2 4
7 9 1 1
3 4 2 3
4 5 1 3
outputCopy
10011
Note
In the first query we can achieve the result, for instance, by using transitions .

The third query asks for changing AAB to A — but in this case we are not able to get rid of the character ‘B’.

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analysis

• 大结论题

• 可以这么想$B\to AC\to AAB\to AAAC \to C$，$C$也同理可以变成$B$

• 就是说$B,C$没有区别可以互换，于是我们把$A$看成$0$，把$B,C$看成$1$

• 接下来要认真考虑各种情况

• 首先除了末尾连续的$0$不能消去，字符串中间的任何$0$我们可以用$1$凭空变出三个$0$来消掉

• 消完后设$S$串中有$S_0$个$0$、$S_1$个$1$，$T$串中有$T_0$个$0$、$T_1$个$1$，这些东西用前缀和搞就好了

• 明显变换过程中，$S_1$只会每次$+2$或者不变，所以$S_1<T_1$且$S_1≡T_1(mod2)$才有解，否则无解

• 当$S_0<T_0$无解，因为$S_0$要么不变要么每次$-2$，不可能变多

• 当$S_1=0≠T_1$时，如果$S_0=T_0$无解，因为想变出两个$1$必须用$0$

• 当$S_1<T_1$时有解，可以把$0$变成$11$再变出$0$来消$0$，否则还要$S_0≡T_0(mod3)$

• 大概就这样了吧……

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code

#include<stdio.h>
#define MAXN 100010
#define min(x,y) (x<y)?(x):(y)

using namespace std;

char st1[MAXN],st2[MAXN];
int a1[MAXN],a2[MAXN],b1[MAXN],b2[MAXN];
int n;

int read()
{
    int x=0,f=1;
    char ch=getchar();
    while (ch<'0' || '9'<ch)
    {
        if (ch=='-')f=-1;
        ch=getchar();   
    }
    while ('0'<=ch && ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

int main()
{
    //freopen("readin.txt","r",stdin);
    scanf("%s%s",st1+1,st2+1);
    for (int i=1;st1[i];i++)
    {
        if (st1[i]=='A')a1[i]=a1[i-1]+1;
        b1[i]=b1[i-1]+(st1[i]!='A');
    }
    for (int i=1;st2[i];i++)
    {
        if (st2[i]=='A')a2[i]=a2[i-1]+1;
        b2[i]=b2[i-1]+(st2[i]!='A');
    }
    n=read();
    while (n--)
    {
        int l1=read(),r1=read(),l2=read(),r2=read(),ok=1;
        int s1=b1[r1]-b1[l1-1],s2=b2[r2]-b2[l2-1];
        int x1=min(r1-l1+1,a1[r1]),x2=min(r2-l2+1,a2[r2]);
        printf("%d",(s1>s2 || (s1+s2)%2==1 || x1<x2 || s1==s2 && x1%3!=x2%3 || s1==0 && x1==x2 && s1!=s2)?0:1);
    }
    return 0;
}