UVA10025 - The ? 1 ? 2 ? ... ? n = k problem

最新推荐文章于 2021-03-04 14:56:10 发布

转载最新推荐文章于 2021-03-04 14:56:10 发布 · 95 阅读

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原文链接：http://www.cnblogs.com/terryX/archive/2013/03/02/2939773.html

本文探讨了如何通过算法和公式解决特定数学问题，详细解释了求解过程，并提供了实例演示。

算法需要自己猜，参考了题解。

题目：

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

Alex Gevak
September 15, 2000 (Revised 4-10-00, Antonio Sanchez)

解答：

 1 #include<stdio.h>
 2 int main()
 3 {
 4     int i,j,t;
 5     scanf("%d",&t);
 6     while(t--)
 7     {
 8         int k;
 9         scanf("%d",&k);
10         if(k<0)
11             k=-k;
12         for(i=1;;i++)
13         {
14             if((1+i)*i/2>=k)
15             {
16                 int flag=1,res=(1+i)*i/2;
17                 for(j=i;flag;)
18                 {
19                     if((res-k)%2==0)
20                         flag=0;
21                     else
22                     {
23                         j++;
24                         res+=j;
25                     }
26                 }
27                 if(flag==0)
28                     break;
29             }
30         }
31         printf("%d\n",j);
32         if(t)
33             printf("\n");
34     }
35     return 0;
36 }

转载于:https://www.cnblogs.com/terryX/archive/2013/03/02/2939773.html