OpenJudge/Poj 1844 Sum

1.链接地址：

http://bailian.openjudge.cn/practice/1844

http://poj.org/problem?id=1844

2.题目：

Sum

Time Limit: 1000MS		Memory Limit: 30000K
Total Submissions: 10031		Accepted: 6564

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

12

Sample Output

7

Hint

The sum 12 can be obtained from at least 7 terms in the following way: 12 = -1+2+3+4+5+6-7.

Source

Romania OI 2002

3.思路：

4.代码：

 1 #include "stdio.h"
 2 //#include "stdlib.h"
 3 int main()
 4 {
 5     int m;
 6     scanf("%d",&m);
 7     int k = (int)((sqrt((float)(1)+8*m)+1)/2);
 8     while((k*k+k-2*m)%4!=0)
 9     {
10         k++;
11     }
12     printf("%d",k);
13     //system("pause");
14     return 0;
15 }

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OpenJudge/Poj 1844 Sum

原文：http://www.cnblogs.com/mobileliker/p/3572429.html