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

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 }

 

你可能感兴趣的:(open)