Writing Code - CodeForces 543 A dp

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本文探讨了在大型项目中，如何合理分配不同能力程序员的代码编写任务，以确保总Bug数量不超过限定值。通过动态规划算法，计算出在特定条件下，有效的任务分配方案数量。

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

题意：有n个人敲m行代码，每个人每行都会出现固定数目的bug，要求bug数不超过b，求有多少种方式。

思路：先循环n个人，dp[i][j]表示敲了i行，出现j个bug的情况数，每次加上这个人的一行，一直递推到最后。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
int T,t,n,m,K,num[510];
ll dp[1010][1010],MOD,ans;
int main()
{
    int i,j,k,a,b,p;
    scanf("%d%d%d%I64d",&n,&m,&K,&MOD);
    for(i=1;i<=n;i++)
       scanf("%d",&num[i]);

    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    for(i=1;i<=n;i++)
       for(j=0;j<m;j++)
          for(k=0;k<=K-num[i];k++)
             dp[j+1][k+num[i]]=(dp[j+1][k+num[i]]+dp[j][k])%MOD;
    for(k=0;k<=K;k++)
       ans+=dp[m][k];
    ans%=MOD;
    printf("%I64d\n",ans);
}