Writing Code（DP）

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

n个人，m行代码，最多有b个错误，每个人写一行有a[i]个bug，问满足的分配方案的个数。

写了三遍总算是会了（……），dp[i][j][k]，前i个人，写了j行，小于等于k个bug的方案数。
然后转移方程就是
dp[i][j][k] = dp[i - 1][j][k] + dp[i][j - 1][k - a[i]];
意思就是这个状态是由第i个人不写第j行和写第j行转移过来。
此时发现第一维可以省略掉（不开这个空间）

#include<cstdio>
#include<cstring>
#include<iostream>
#include<queue>
#include<vector>
#include<algorithm>
#include<string>
#include<cmath>
#include<set>
#include<map>
#include<vector>
using namespace std;
typedef long long ll;
const int inf = 0x3f3f3f3f;
const int maxn = 1005;
int dp[505][505];
int a[505];
int main()
{
    #ifdef LOCAL
    freopen("C:\\Users\\巍巍\\Desktop\\in.txt","r",stdin);
    //freopen("C:\\Users\\巍巍\\Desktop\\out.txt","w",stdout);
    #endif // LOCAL
    int n,m,b,mod;
    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 = 0;k <= b;k++)
            {
                if(k < a[i])dp[j][k] = dp[j][k];
                else
                    dp[j][k] = (dp[j][k] + dp[j - 1][k - a[i]])%mod;
            }
        }
    }
    printf("%d\n",dp[m][b]);
    return 0;
}