Consecutive Sum Riddle (CodeForces - 1594A)

原创已于 2023-12-07 02:06:36 修改 · 960 阅读

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#c++ #算法 #数据结构

于 2023-12-07 00:31:13 首次发布

探讨如何使用编程解决一道关于连续整数和的题目，适用于CodeForces1594A，涉及输入整数n并找到满足条件的l和r。

第一周题单

B - Consecutive Sum Riddle (CodeForces - 1594A)

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 n. You need to find two integers l and r such that $10^{18}≤l<r≤10^{18}$ and $l + (l + 1) + \dots + (r - 1) + r = n$ .

Input

The first line contains a single integer $t(1≤t≤10^4)$ — the number of test cases.
The first and only line of each test case contains a single integer $n(1≤n≤10^{18})$ .

Output

For each test case, print the two integers l and r such that $10^{18}≤l<r≤10^{18}$ and $l + (l + 1) + \dots + (r - 1) + r = n$ .
It can be proven that an answer always exists. If there are multiple answers, print any.

Simple1

Input

7
1
2
3
6
100
25
3000000000000

Output

0 1
-1 2
1 2
1 3
18 22
-2 7
999999999999 1000000000001

Note

In the first test case, 0+1=1.
In the second test case, (−1)+0+1+2=2.
In the fourth test case, 1+2+3=6.
In the fifth test case, 18+19+20+21+22=100.
In the sixth test case, (−2)+(−1)+0+1+2+3+4+5+6+7=25.

题目大意

找到一串从 $l$ 到 $r$ 的数列，使数列之和为 $n$

题解过程

$∑i=−n+1n=n\sum_{i=-n+1}^{n}=n$

代码部分(cpp)

#include <iostream>
using namespace std;
void solution()
{
    int t;
    cin >> t;
    while (t--)
    {
        long long num;
        cin >> num;
        cout << -(num - 1) << " " << num << endl;
    }
}
int main(int argc, const char **argv)
{
    solution();
    return 0;
}