poj 1844 Sum

最新推荐文章于 2021-01-30 20:25:47 发布

原创最新推荐文章于 2021-01-30 20:25:47 发布 · 448 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#numbers #output #integer #input #each

简单题专栏收录该内容

4 篇文章

订阅专栏

本文探讨了一道数学问题：通过给定的整数S，找到最小的自然数N，使得从1到N的所有整数之和能够通过调整每个数的正负号达到S。文章提供了一种有效的算法实现，并附带源代码。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

Consider the natural numbers from 1 to N. By associating to each number a sign (+ or -) and calculating the value of this expression we obtain a sum S. The problem is to determine for a given sum S the minimum number N for which we can obtain S by associating signs for all numbers between 1 to N.

For a given S, find out the minimum value N in order to obtain S according to the conditions of the problem.

Input

The only line contains in the first line a positive integer S (0< S <= 100000) which represents the sum to be obtained.

Output

The output will contain the minimum number N for which the sum S can be obtained.

Sample Input

Sample Output

这道题乍看一下有点摸不着头脑~

但实际就是求sum=1+2+3+4+……+n~

当sum除以input的余数为偶数时~n为所求位数~

因为若余2*k~

当k=1时~将+1改为-1就行~

当k=2时~将+2改为-2就行~以此类推~

当余数为奇数时~则只需继续循环~加上n+1~这样余数就为偶数了~

下面是代码~

#include"stdio.h"
int main()
{
int s;
int n;
int sum;
while(scanf("%d",&s)!=EOF)
{
 sum=1;
 n=1;
 while((sum-s)%2==1||sum<s)
 {
 n++;
 sum+=n;

 }
 printf("%d\n",n);
}
return 0;
}