查询范围[L，R]

Question

Given a binary string (that is a string consisting of only 0 and 1). They were supposed to perform two types of query on the string. Problem

Type 0: Given two indices l and r.Print the value of the binary string from l to r modulo 3.

Type 1: Given an index l flip the value of that index if and only if the value at that index is 0.

I am trying to solve this using BIT .

If the number in range [l,r] is even then:
if the sum of the numbers of one is even then the answer is 0 else 2

If the number in range [l,r] is odd
if the sum of the numbers of one is even then the answer is 0 else 1

But I am getting wrong answer for some test cases what wrong is in my approach.

public static void update(int i){

    while(A.length>i){
        A[i]+=1;
        i+=i&-i;
    }

}

public static int ans(int i){
    int a=0;

    while(i>0){
        a+=A[i];
        i-=i&-i;
    }
    return a;
}

Answer for each Query.

while(Q>0){
    Q--;
    int x = in.nextInt();
    int l = in.nextInt()+1;
    if(x==1){
        if((ans(l)-ans(l-1))==0) update(l);

        continue;
    }
    int r  = in.nextInt()+1;

    int f = ans(r) - ans(r-1);

    if(f==0){

        int sum = ans(r)- ans(l-1);
        if(sum%2==0) System.out.println(0);
        else System.out.println(2);
    }else{

        int sum = ans(r)- ans(l-1);
        if(sum%2==0) System.out.println(0);
        else System.out.println(1);

    }
}

Full CODE

Answer 1

When building a binary number, from the left digit to the right digit, if you consider only a currently parsed section of the string, it is a binary number. We'll call it n.

When you append a digit to the right of n, this shifts it left, and then also adds the digit (1 or 0). So n0 = 2*n and n1 = 2*n+1

Because you only care about the number modulo 3, you can just keep track of this modulo 3.

You can note that

0 (mod 3) * 2     = 0 (mod 3)
0 (mod 3) * 2 + 1 = 1 (mod 3)
1 (mod 3) * 2     = 2 (mod 3)
1 (mod 3) * 2 + 1 = 0 (mod 3)
2 (mod 3) * 2     = 1 (mod 3)
2 (mod 3) * 2 + 1 = 2 (mod 3)

You can construct a simple fsm to represent these relationships and simply use the section of the string which you are interested in as input to it. Or implement it however else you want.

Hopefully you realise that "Given an index l flip the value of that index if and only if the value at that index is 0." simply means set the value at l to 1.