Codeforces 544C Writing Code【二维完全背包】

最新推荐文章于 2021-05-23 19:02:44 发布

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最新推荐文章于 2021-05-23 19:02:44 发布

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分类专栏：水题背包文章标签： Codeforces 544C

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本文探讨了一组程序员如何在限定的bug数量内完成指定规模的编程任务。通过使用完全背包算法，文章提供了一种有效的方法来计算不同人员配置下可能的代码编写方案总数。

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Programmers working on a large project have just received a task to write exactly m lines of code. There are n programmers working on a project, the i-th of them makes exactly a_i bugs in every line of code that he writes.

Let's call a sequence of non-negative integers v₁, v₂, ..., v_n a plan, if v₁ + v₂ + ... + v_n = m. The programmers follow the plan like that: in the beginning the first programmer writes the first v₁ lines of the given task, then the second programmer writes v₂ more lines of the given task, and so on. In the end, the last programmer writes the remaining lines of the code. Let's call a plan good, if all the written lines of the task contain at most b bugs in total.

Your task is to determine how many distinct good plans are there. As the number of plans can be large, print the remainder of this number modulo given positive integer mod.

Input

The first line contains four integers n, m, b, mod (1 ≤ n, m ≤ 500, 0 ≤ b ≤ 500; 1 ≤ mod ≤ 10⁹ + 7) — the number of programmers, the number of lines of code in the task, the maximum total number of bugs respectively and the modulo you should use when printing the answer.

The next line contains n space-separated integers a₁, a₂, ..., a_n (0 ≤ a_i ≤ 500) — the number of bugs per line for each programmer.

Output

Print a single integer — the answer to the problem modulo mod.

Examples

Input

3 3 3 100
1 1 1

Output

Input

3 6 5 1000000007
1 2 3

Output

Input

3 5 6 11
1 2 1

Output

题目大意：

现在一共有N个程序员，一个程序员（编号为i）写一行代码会有Ai个bug.目的是完成一个M行的程序。

第i个程序员会写Vi行代码，对应就会出现Vi*Ai个Bug.

问将这个M行的程序写完之后，Bug数量不超过b个的分配方案数。

思路：

题目读懂之后，很显然的一个完全背包。

设定dp【i】【j】表示一共写了i行代码，一共写出来了j个Bug的方案数。

那么很简单的递推方程：dp【i】【j】=dp【i-1】【j-Ai】；

那么ans=Σdp【m】【i】（0<=i<=b）

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int dp[505][500*4];
int main()
{
    int n,m,b,mod;
    while(~scanf("%d%d%d%d",&n,&m,&b,&mod))
    {
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int z=0;z<n;z++)
        {
            int x;
            scanf("%d",&x);
            for(int i=1;i<=m;i++)
            {
                for(int j=x;j<=b;j++)
                {
                dp[i][j]+=dp[i-1][j-x];
                   dp[i][j]%=mod;
                }
            }
        }
        int ans=0;
        for(int i=0;i<=b;i++)
        {
            ans+=dp[m][i];
            ans%=mod;
        }
        printf("%d\n",ans);
    }
}