POJ 1032 Parliament

最新推荐文章于 2019-07-17 16:24:23 发布

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#ACM #数论

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本文介绍了一个数学问题的解决方法，即如何将指定数量的代表分成不同大小的小组，以确保议会能尽可能长时间地运行。通过将代表拆分为以2开头的连续数列，并对剩余数进行补充的方法，可以最大化议会的工作时间。

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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

Sample Output

3 4

思路：就是说拆分一个整数，使得拆开的数的乘积最大。
考虑一个数，只要能拆成两个大于1的数，则乘积一定大于等于原来的数。所以我们希望将它全部拆成2（或者n个2和1个3），但因为数字不能重复，所以对任意一个数，我们希望将它拆成以2开头的连续数列，最后剩余的数从大到小补到前面。

代码：

#include <cstdio>
#include <iostream>
#include <cstring>
#include <vector>
#include <map>
#include <string>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <cstdlib>

using namespace std;

int main(void)
{
    #ifdef LOCAL
        freopen("in.txt","rb",stdin);
        //freopen("out.txt","wb",stdout);
    #endif // LOCAL
    int n,s[1000];
    scanf("%d",&n);
    int l;
    for(l=2;n>=l;l++)
    {
        n-=l;
        s[l]=l;
    }
    l--;
    for(int i=2;i<=l;i++)
        s[i]+=n/(l-2+1);
    n=n%(l-2+1);
    for(int i=l;i>=l-n+1;i--)
        s[i]+=1;
    for(int i=2;i<=l;i++)
    {
        cout << s[i];
        if(i!=l) cout << ' ';
    }
    return 0;
}