you will get TLE if you try to use Loop for all numbers
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jacob answer
Let's introduce
f(x) = x ^ (x-1) ^ (x-2) ^ ... ^ 1
Then anwer would be
f(end) ^ f(start - 1)
Let's now learn to calculate f(x) fast enough. First, notice that f(4*k - 1) = 0 (provable by induction). So to calculate f(x) you only need to take at most 4 last numbers and xor them.
Finally,
f(4*k) = 4*k
f(4*k + 1) = 1
f(4*k + 2) = 4*k + 3
f(4*k + 3) = 0
Expressions "at most 4 last numbers" and "only last 4 numbers" are different. In fact, you need to take exactly (x+1)%4 last numbers.
Conclusion:
Use this function
int f(int n){ switch(n%4){ case 0: return n; case 1: return 1; case 2: return n+1; default: return 0; } }so if you want to get Xor between numbers in range start to end
f(start)^f(end-1)