Codeforces Round #302 (Div. 1) 543A Writing Code(dp)

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本文探讨了如何在大型项目中，通过合理分配程序员的工作量和任务，确保编写出的代码总bug数量不超过设定阈值的最优策略。通过完全背包问题的解决思路，实现代码质量和效率的双提升。

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个程序员写m行代码，一行会出现a[i]个bug，问你出现bug总数不超过b的方案数。

完全背包的思想，将dp降维优化，状态转移方程：dp[k][j] += dp[k - 1][j - a[i]]。

AC代码：

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
using namespace std;
const int MAXN = 510;
typedef long long ll;
int n, m, b, mod, a[MAXN];
ll dp[MAXN][MAXN];
int main(int argc, char const *argv[])
{
	scanf("%d%d%d%d", &n, &m, &b, &mod);
	for(int i = 1; i <= n; ++i)
		scanf("%d", &a[i]);
	dp[0][0] = 1;
	for(int i = 1; i <= n; ++i)
		for(int k = 1; k <= m; ++k)
			for(int j = a[i]; j <= b; ++j) {
				dp[k][j] += dp[k - 1][j - a[i]];
				dp[k][j] %= mod;
			}
	ll ans = 0;
	for(int i = 0; i <= b; ++i)
		ans += dp[m][i];
	printf("%lld\n", ans % mod);
	return 0;
}