UVa 10025: The ? 1 ? 2 ? ... ? n = k problem

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