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

標籤：

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個人寫m行代碼，最大的bug錯誤為b，模數為mod，下面n行表示這n個人寫一行代碼為犯得錯誤，問你一共有多少種情況是的bug數目不超過b

定義 dp[i][j] 表示選定了i個人，犯得bug為j的種類   那麼得到狀態轉移方程 dp[j][k] =(dp[j][k] + dp[j-1][k-a[i]])%mod

表示當前j個人k個bug可以由原來的以及少一個人之後選擇那個人轉移過來

注意初始化為dp[0][0...b] = 1 表示不選擇人的時候所有的bug的種類犯錯誤都只有一種

#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\n",dp[m][b]);    }    return 0;}

　　

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