Contents
Consecutive Sum Riddle codeforces solution

For Solution
Click Here!
Theofanis has a riddle for you and if you manage to solve it, he will give you a Cypriot snack halloumi for free (Cypriot cheese).
You are given an integer nn. You need to find two integers ll and rr such that −1018≤l<r≤1018−1018≤l<r≤1018 and l+(l+1)+…+(r−1)+r=nl+(l+1)+…+(r−1)+r=n.
Consecutive Sum Riddle solution codeforces
The first line contains a single integer tt (1≤t≤1041≤t≤104) — the number of test cases.
The first and only line of each test case contains a single integer nn (1≤n≤10181≤n≤1018).
Consecutive Sum Riddle solution codeforces
For each test case, print the two integers ll and rr such that −1018≤l<r≤1018−1018≤l<r≤1018 and l+(l+1)+…+(r−1)+r=nl+(l+1)+…+(r−1)+r=n.
It can be proven that an answer always exists. If there are multiple answers, print any.
input
Consecutive Sum Riddle solution codeforces
7 1 2 3 6 100 25 3000000000000
Consecutive Sum Riddle solution codeforces
0 1 1 2 1 2 1 3 18 22 2 7 999999999999 1000000000001
Consecutive Sum Riddle solution codeforces
In the first test case, 0+1=10+1=1.
In the second test case, (−1)+0+1+2=2(−1)+0+1+2=2.
In the fourth test case, 1+2+3=61+2+3=6.
In the fifth test case, 18+19+20+21+22=10018+19+20+21+22=100.
In the sixth test case, (−2)+(−1)+0+1+2+3+4+5+6+7=25(−2)+(−1)+0+1+2+3+4+5+6+7=25.

For Solution
Click Here!