Codeforces 544C Writing Code【二维完全背包】

C. Writing Code

time limit per test
3 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

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.

Examples

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

题目大意：

现在一共有N个程序员，一个程序员（编号为i）写一行代码会有Ai个bug.目的是完成一个M行的程序。

第i个程序员会写Vi行代码，对应就会出现Vi*Ai个Bug.

问将这个M行的程序写完之后，Bug数量不超过b个的分配方案数。

思路：

题目读懂之后，很显然的一个完全背包。

设定dp【i】【j】表示一共写了i行代码，一共写出来了j个Bug的方案数。

那么很简单的递推方程：dp【i】【j】=dp【i-1】【j-Ai】；

那么ans=Σdp【m】【i】（0<=i<=b）

Ac代码：

#include<stdio.h>
#include<string.h>
using namespace std;
int dp[505][500*4];
int main()
{
int n,m,b,mod;
while(~scanf("%d%d%d%d",&n,&m,&b,&mod))
{
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int z=0;z<n;z++)
{
int x;
scanf("%d",&x);
for(int i=1;i<=m;i++)
{
for(int j=x;j<=b;j++)
{
dp[i][j]+=dp[i-1][j-x];
dp[i][j]%=mod;
}
}
}
int ans=0;
for(int i=0;i<=b;i++)
{
ans+=dp[m][i];
ans%=mod;
}
printf("%d\n",ans);
}
}