Valera and Contest（模拟）

 

B. Valera and Contest

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Valera loves to participate in competitions. Especially in programming contests. Today he has participated in the contest with his team, consisting of n students (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 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, Valera did not find out exactly how many points each of n students 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 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 guaranteed that the input is such that the answer exists.

Output

Print exactly n integers a₁, a₂, ..., a_n — 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 

 

   题意：

   给出n,k,l,r,sall,sk。求一序列满足条件：

   1、该序列有n个数字组成，且序列总和等于sall；

   2、该序列的n个数均由l到r的数字组成；

   3、这个序列降序排序后，前k个数和为sk；

   满足序列可能有多个，输出其中一个即可。

 

   思路：

   1.先将前k个数赋值为r，若k*r>sk，则逐一减一直到等于sk；

   2.剩下的n-k个数赋值为l，若(n-k)*l<sall，则逐一加一直到等于sall-sk；

 

   AC：

#include<stdio.h>
int num[1005];

int main()
{
    int n,k,l,r,sa,sk,sum,ans;
    scanf("%d%d%d%d%d%d",&n,&k,&l,&r,&sa,&sk);
    for(int i=1;i<=n;i++)
    {
       if(i<=k) num[i]=r;
       if(i>k)  num[i]=l;
    }
    sum=r*k;
    ans=1;
    while(sum!=sk)
    {
        if(ans==(k+1)) ans=1;
        num[ans++]--;
        sum--;
    }
    sum=(n-k)*l;
    ans=k+1;
    while(sum!=(sa-sk))
    {
        if(ans==(n+1)) ans=k+1;
        num[ans++]++;
        sum++;
    }
    for(int i=1;i<=n;i++)
    {
        printf("%d",num[i]);
        i==n?printf("\n"):printf(" ");
    }
    return 0;
}