这道题仔细思考后就可以得到比较快捷的解法，只要求出满足n*(n+1)/2 >= |k| ，且n*(n+1)/2-k为偶数的n就可以了。注意n==0时需要特殊判断。
我的解题代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <string>
#include <algorithm>
using namespace std;
int main()
{
long long T,K,k,n;
cin >> T;
while(T--)
{
cin >> K; if(K<0) k=-K; else k=K;
if(K==0) cout << 3 << endl;
else
{
double ans = (sqrt(1.0+8*k)-1)/2;
n=ceil(ans);
while((n*(n+1)/2-K)%2) n++;
cout << n << endl;
}
if(T) cout << endl;
}
return 0;
}
附上题目如下：
The problem
Given the following formula, one can set operators '+' or '-' instead of each '?', in order to obtain a given k
? 1 ? 2 ? ... ? n = k
For example: to obtain k = 12 , the expression to be used will be:
- 1 + 2 + 3 + 4 + 5 + 6 - 7 = 12
with n = 7
The Input
The first line is the number of test cases, followed by a blank line.
Each test case of the input contains integer k (0<=|k|<=1000000000).
Each test case will be separated by a single line.
The Output
For each test case, your program should print the minimal possible n (1<=n) to obtain k with the above formula.
Print a blank line between the outputs for two consecutive test cases.
Sample Input
2 12 -3646397
Sample Output
7 2701