CodeForces 369B Valera and Contest

最新推荐文章于 2023-04-02 21:00:38 发布

最新推荐文章于 2023-04-02 21:00:38 发布 · 107 阅读

在一次编程竞赛中，团队成员各自获得的成绩范围为l到r分，总分为sall分，最高分数之和为sk分。任务是找出一种可能的成绩分配方案，满足所有条件。

Valera loves to participate in competitions. Especially in programming contests. Today he has participated in the contest with his team, consisting ofnstudents (including Valera). This contest was an individual competition, so each student in the team solved problems individually.

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

each student in the team scored at leastlpoints and at mostrpoints;
in total, all members of the team scored exactlysallpoints;
the total score of thekmembers of the team who scored the most points is equal to exactlysk; more formally, ifa1, a2, ..., anis the sequence of points earned by the team of students in the non-increasing order(a1 ≥ a2 ≥ ... ≥ an), thensk = a1 + a2 + ... + ak.

However, Valera did not find out exactly how many points each ofnstudents scored. Valera asked you to recover any distribution of scores between the students of the team, such that all the conditions above are met.

Input

The first line of the input contains exactly six integersn, k, l, r, sall, sk(1 ≤ n, k, l, r ≤ 1000;l ≤ r;k ≤ n;1 ≤ sk ≤ sall ≤ 106).

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

Output

Print exactlynintegersa1, a2, ..., an— the number of points each student scored. If there are multiple solutions, you can print any of them. You can print the distribution of points in any order.

Sample test(s)

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 

一定要注意 n==k 不然会RE的

#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;
}