Codeforces Round #302 (Div. 2).C. Writing Code (dp)-优快云博客

本文探讨了在大型项目中，程序员遵循特定计划完成指定代码行数任务时，如何通过分配代码行数来最小化总错误数量，同时不超过设定的错误上限。文章详细介绍了输入参数、代码错误计算、输出结果以及使用记忆化搜索算法解决此问题的方法。

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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 a_i bugs in every line of code that he writes.

Let's call a sequence of non-negative integers v₁, v₂, ..., v_n a plan, if v₁ + v₂ + ... + v_n = 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, ..., a_n (0 ≤ a_i ≤ 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

input

3 6 5 1000000007 1 2 3

output

input

3 5 6 11 1 2 1

output

 1 #include<stdio.h>
 2 #include<string.h>
 3 const int M = 505 ;
 4 int dp[M][M] ;
 5 int a[M] ;
 6 int n , m , b , mod ;
 7 
 8 int main ()
 9 {
10    // freopen ("a.txt" , "r" , stdin ) ;
11     while (~ scanf ("%d%d%d%d" , &n , &m , &b , &mod )) {
12         memset (dp , 0 , sizeof(dp)) ;
13         for (int i = 0 ; i < n ; i ++) scanf ("%d" , &a[i]) ;
14         int sum = 0 ;
15         memset (dp , 0 , sizeof(dp)) ;
16         for (int i = 0 ; i <= m ; i ++) {
17             int bug = i * a[0] ;
18             if (bug <= b) dp[i][bug] = 1 ;
19         }
20         for (int i = 1 ; i < n ; i ++) {
21             for (int j = 0 ; j < m ; j ++) {
22                 for (int k = 0 ; k <= b ; k ++) {
23                     if (dp[j][k]) {
24                         int x = j + 1 , y = k + a[i] ;
25                         if (y <= b) {
26                           //  printf ("(%d , %d) = %d ---> (%d , %d) = %d\n" , j , k , dp[j][k] , x , y , dp[x][y]) ;
27                             dp[x][y] += dp[j][k] ;
28                             dp[x][y] %= mod ;
29                         }
30                     }
31                 }
32             }
33         }
34         for (int i = 0 ; i <= b ; i ++) {
35             sum += dp[m][i] ;
36             sum %= mod ;
37         }
38         printf ("%d\n" , sum ) ;
39     }
40 }