CodeForces 369B

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本文针对CodeForces369B比赛题目进行了详细的解析，包括题目的核心要求、解决思路及代码实现。重点在于如何分配团队成员得分以符合给定条件。

CodeForces 369B

Description

Valera loves toparticipate in competitions. Especially in programming contests. Today he hasparticipated in the contest with his team, consisting of n students(including Valera). This contest was an individual competition, so each studentin the team solved problems individually.

After thecontest was over, Valera was interested in results. He found out that:

each student in the team scored at least l points and at most r points;
in total, all members of the team scored exactly s_all points;
the total score of the k members of the team who scored the most points is equal to exactly s_k; more formally, if a₁, a₂, ..., a_n is the sequence of points earned by the team of students in the non-increasing order (a₁ ≥ a₂ ≥ ... ≥ a_n), then s_k = a₁ + a₂ + ... + a_k.

However, Valeradid not find out exactly how many points each of n studentsscored. Valera asked you to recover any distribution of scores between thestudents of the team, such that all the conditions above are met.

Input

The first lineof the input contains exactly six integers n, k, l, r, s_all, s_k (1 ≤ n, k, l, r ≤ 1000; l ≤ r; k ≤ n; 1 ≤ s_k ≤ s_all ≤ 10⁶).

It's guaranteedthat the input is such that the answer exists.

Output

Printexactly n integers a₁, a₂, ..., a_n — the numberof points each student scored. If there are multiple solutions, you can printany of them. You can print the distribution of points in any order.

Sample Input

Input

5 3 1 3 13 9

Output

2 3 2 3 3

Input

5 3 1 3 15 9

Output

3 3 3 3 3

一.                     题意分析

有n个人，每个人最少l分，最多r分，所有人 s_all分，最多的k个人s_k分。输出这些人的分数。

二.                     思路过程

先分析k个人得s_k分:

sk%k == sk-sk/k*k

也就是说sk%k个人是分数是sk/k+1的

输出sk%k个sk/k+1

然后有k个人的分数是sk/k分

输出k个sk/k

然后把s_all减去s_k分，得到的就是剩下来n-k个人的分数，然后进行分析:如果n-k是0了，那么说明这时人已经全部输出了；如果n-k不是0，那么说明此时还没有全部输出。同样的，有s_all-s_k分，有n-k个人，计算方法和上面一样。

三.代码

<pre name="code" class="cpp"><span style="font-family:Courier New;font-size:18px;">#include <iostream>
#include <cstring>
#include <cstdio>

using namespace std;

const int maxn=1100;

int  sorce[maxn],n,k,l,r,sall,sk;

int main()
{
    cin>>n>>k>>l>>r>>sall>>sk;
    for(int i=1;i<=k;i++)
    {
        sorce[i]=sk/k;
    }
    int t=sk%k,i=1;
    while(t)
    {
        t--; sorce[i++]++;
    }
    if(k!=n)
    {
        for(int i=k+1;i<=n;i++)
        {
            sorce[i]=(sall-sk)/(n-k);
        }
        int t,i;
        t=(sall-sk)%(n-k),i=k+1;
        while(t)
        {
            t--; sorce[i++]++;
        }
    }
    for(int i=1;i<=n;i++)
    {
        cout<<sorce[i]<<" ";
    }
    return 0;
}
</span>