Parliament

Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 18707   Accepted: 7941

Description

New convocation of The Fool Land's Parliament consists of N delegates. According to the present regulation delegates should be divided into disjoint groups of different sizes and every day each group has to send one delegate to the conciliatory committee. The composition of the conciliatory committee should be different each day. The Parliament works only while this can be accomplished.
You are to write a program that will determine how many delegates should contain each group in order for Parliament to work as long as possible.

Input

The input file contains a single integer N (5<=N<=1000 ).

Output

Write to the output file the sizes of groups that allow the Parliament to work for the maximal possible time. These sizes should be printed on a single line in ascending order and should be separated by spaces.

Sample Input

7

Sample Output

3 4

Source

Northeastern Europe 1998
 
分析：
题目的意思是，将一个数字拆分成互不相同的数，使它们的积最大。
 
解题：
如果不考虑条件“互不相同的”，那么最终结果肯定是2\3的组合（4拆2+2效果一样），比如10=5+5=2+3+2+3。
将“互不相同”纳入考虑范围，那么，这个组合必然是从2开始的，尽可能小的，一组数列。
那么可以，从2+3+4+...加到一个不大于N的值，再将剩余的值从大到小加回去。比如，26=2+3+4+5+6....6，再将6从大到小依次加回去，就得到答案3+4+5+6+8
 
 1 #include <stdio.h>
 2 
 3 int main(void)
 4 {
 5     int ans[100] = {0};
 6     int all = 0;
 7     int sum = 0;
 8     int end = 2;
 9     int num = 0;
10     int j = 0;
11     int remaind = 0;
12     scanf("%d",&all);
13 
14     while(1)
15     {
16         if(sum+end <= all)
17         {
18             sum = sum+end;
19             ans[num] = end;
20             num++;
21             end++;
22         }
23         else
24         {
25             remaind = all - sum;
26             break;
27         }
28     }
29 
30     j = num - 1;
31     while(remaind > 0)
32     {
33         if(j<0) j=num-1;
34         ans[j]++;
35         j--;
36         remaind--;
37     }
38 
39     for(j=0; j<num; j++)
40     {
41         printf("%d ", ans[j]);
42     }
43 
44     return 0;
45 }