Codeforces Round #302 (Div. 2)C (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.

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

把m写成了n，dp一直调不出来，囧rz,dp[i][j]表示前i天bug为j的方案数，希望今后比赛能细心点。

#include <bits/stdc++.h>
#define foreach(it,v) for(__typeof((v).begin()) it = (v).begin(); it != (v).end(); ++it)
using namespace std;
typedef long long ll;
const int maxn = 5e2 + 5;
int a[maxn],mod,b,n,m;
ll dp[maxn][maxn];
int main()
{
while(cin>>n>>m>>b>>mod) {
memset(dp,0,sizeof dp);
for(int i = 1; i <= n; i++)cin>>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 res = 0;
for(int i = 0; i <= b; i++)res += dp[m][i];
cout<<res%mod<<endl;
}
return 0;
}