C. Writing Code（Codeforces Round #302（div2）

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本文探讨了在大型项目中，n位程序员如何通过制定合理的计划来高效完成编写m行代码的任务，同时确保代码总bug数量不超过b个，通过使用mod运算计算不同可行计划的数量。该文提供了一个算法解决方案，并通过实例演示了实现过程。

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 ai bugs in every line of code that he writes.

Let's call a sequence of non-negative integers v1, v2, ..., vn a plan, if v1 + v2 + ... + vn = 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, ..., an (0 ≤ ai ≤ 500) — the number of bugs per line for each programmer.

Output

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

Sample test(s)

input
3 3 3 100
1 1 1

output
10

input
3 6 5 1000000007
1 2 3

output
0

input
3 5 6 11
1 2 1

output
0






题目大意：


    一段程序m段，n个程序员，给出每个程序员写段代码的bugs数，每个程序员都是写的连续的段，求有多少


种方法。




题解：
  
    dp[j][k],i是选择的代码段，k是选择的bugs数，由于是连续的，所以最外层的肯定是每个程序的bugs


数，再枚举j，k就好了，dp[j][k]=(dp[j][k]+dp[j-1][k-v[i]])%mod。






代码：


#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int maxn=505;
int dp[maxn][maxn];
int v[maxn];
int main()
{
    int m,n,b,mod;
    while(~scanf("%d%d%d%d",&n,&m,&b,&mod))
    {
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&v[i]);
        }
        memset(dp,0,sizeof(dp));
        dp[0][0]=1;
        for(int i=1;i<=n;i++)
        {
            for(int j=1;j<=m;j++)
            {
                for(int k=v[i];k<=b;k++)
                {
                    dp[j][k]=(dp[j][k]+dp[j-1][k-v[i]])%mod;
                }
            }
        }
        int ans=0;
        for(int i=0;i<=b;i++)
        ans=(ans+dp[m][i])%mod;
        printf("%d\n",ans);
    }
    return 0;
}