Codeforces Round #302 (Div. 2)——C dp—— Writing Code

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

 大意：有n个人写m行代码，最大的bug错误为b，取模为mod，下面n行表示这n个人写一行代码为犯得错误，问你一共有多少种情况是的bug数目不超过b

定义 dp[i][j] 表示选定了i个人，犯得bug为j的种类   那么得到状态转移方程 dp[j][k] =(dp[j][k] + dp[j-1][k-a[i]])%mod

表示当前j个人k个bug可以由原来的以及少一个人之后选择那个人转移过来

注意初始化为dp[0][0...b] = 1 表示不选择人的时候所有的bug的种类犯错误都只有一种

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int main()
{
    int dp[550][550];
    int n,m,a[550],mod,b;
    while(~scanf("%d%d%d%d",&n,&m,&b,&mod)){
        for(int i = 1; i <= n ;i++)
            scanf("%d",&a[i]);
        memset(dp,0,sizeof(dp));
        for(int i = 0; i <= b ;i++)
            dp[0][i] = 1;
        for(int i = 1; i <= n ; i++){
            for(int j = 1; j <= m; j++){
                for(int k = a[i]; k <= b ; k++){
                    dp[j][k] = (dp[j][k]+dp[j-1][k-a[i]])%mod;
                }
            }
        }
        printf("%d
",dp[m][b]);
    }
    return 0;
}