# Practice on Toph

Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.

Participate in exhilarating programming contests, solve unique algorithm and data structure challenges and be a part of an awesome community.

Limits
1s, 512 MB

The fight for the best superhero (superman vs batman) in the universal is going to finally come to an end. Expectorates from all across the universal came to witness the breath taking final fight.

The fight began in an astonishing way.A common problem was set for both of them. The problem goes like this.

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.

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.

The problem proved to be a really tough one for both of them.Hours passed by but neither of them could solve the problem.So both of them wants you to solve this problem and then you get the right to choose the best superhero in this universal.

The first line contains N denoting the length of the binary string .The second line contains the N length binary string.Third line contains the integer Q indicating the number of queries to perform.This is followed up by Q lines where each line contains a query.

Constraints

1<= N <=10^5

1<= Q <= 10^5

0 <= l <= r < N

For each query of Type 0 print the value modulo 3.

Input | Output |
---|---|

5 10010 6 0 2 4 0 2 3 1 1 0 0 4 1 1 0 0 3 | 2 1 2 1 |

Query 1 : This is of type 0. The binary string is 010 which is equal to 2 and 2%3=2. So answer is 2.

Query 2 : This is of type 0. The binary string is 01 which is equal to 1 ( (2^1) * 0 +(2^0) * 1 ) =0 + 1 =1) and 1%3=1. So answer is 1.

Query 3 : This is of type 1. The value at index 1 is 0 so we flip it .The new string is 11010.

Query 4 : This is of type 0. The binary string is 11010 ( (2^0) * 0 +(2^1) * 1 +(2^2) * 0 +(2^3) * 1 +(2^4) * 1 = 2 + 8 +16 =26 ) which is equal to 26 and 26%3=2. So answer is 2.

Query 5 : This is of type 1. The value at index 1 is 1 so we do nothing .The new string is 11010.

Query 6 : This is of type 0. The binary string is 1101 ( (2^0) * 1 +(2^1) * 0 +(2^2) * 1 +(2^3) * 1 = 1 + 4+ 8 =13 ) which is equal to 13 and 13%3=1. So answer is 1.

63% Solution Ratio

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