10025 - The ? 1 ? 2 ? ... ? n = k problem-优快云博客

本文链接：https://blog.youkuaiyun.com/sio__five/article/details/9399637

本文探讨了如何通过在数学公式中插入加减操作符，以达到特定数值的目标，并且着重强调了找到最小项数的方法。通过实例演示，详细解释了求解过程和关键步骤。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

The ? 1 ? 2 ? ... ? n = k problem

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

Sample Output

7

2701

（）1（）2（）3（）……（）n=k

要求在括号内填入加号或者减号，使得这n个数能够得到k。求n的最小值

思路：前n个数全部加起来之后，再把其中的几个加号调整为减号，便可以得到k。

那么，必须保证前n个数相加大于等于k。如果等于，就直接输出。如果不等于，那么调整减号，而且必须要注意到，每次调一个减号，减去的是偶数，而且减去的最小值为2

随意要保证前n个数的和的奇偶性和k相同

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main ()
{
    int k,n,i,t,sum;
    cin>>t;
    while(t--)
    {
        cin>>k;
        k=fabs(k);
        n=pow(k,1/2);
        sum=(1+n)*n/2;
        while(sum<k)
        {
            n++;
            sum+=n;
        }
        if (sum==k) printf("%d\n",n);
        else
        {
            if (sum<k+2)
            {
                n++;
                sum+=n;
            }
            while (k%2!=sum%2)
            {
                n++;
                sum+=n;
            }
            printf("%d\n",n);
        }
        if (t!=0) cout<<endl;
    }
    return 0;
}