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最新推荐文章于 2022-05-01 08:41:13 发布

原创最新推荐文章于 2022-05-01 08:41:13 发布 · 445 阅读

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本文介绍了一个程序员团队在完成代码任务时如何通过完全背包算法找出所有可能的计划组合，以确保总错误数量不超过预设限制。文章详细解释了该问题的算法解决思路，并提供了一段示例代码。

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.

Example
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

这道题是完全背包问题，只不过为了不重复，是一件一件物品更新而已，外层枚举个数，内层套一个记录权值的背包，就可以DP了，以后遇到这种取多少个都可以的分层的最大化问题或计数问题，直接上完全背包就可以了

%:pragma GCC optimize(4)
using namespace std;
#include<cstdio>
#include<iostream>
#include<cstring>
#define N 505
long long n,mm,k;
long long mod;
long long dp[N*5][N*5];
long long tot;
long long a[N];
long long m[N*5];
long long v[N*5];
int it[N];
int num=0;
int sum;
int cnt;
int main()
{
    cin>>n>>mm>>k>>mod;
    for(int i=1;i<=n;i++) cin>>a[i];
    it[1]=1;num=1;sum=1;
    for(int i=1;i<=n;i++)
    {
        m[++cnt]=a[i];
        v[cnt]=1;
    }
    dp[0][0]=1;
    for(int i=1;i<=cnt;i++)
    {
        for(int hh=0;hh<=mm-v[i];hh++)
        {
            for(int j=k+1;j>=0;j--)
            {
                dp[hh+v[i]][min(j+m[i],k+1)]+=dp[hh][j];
                dp[hh+v[i]][min(j+m[i],k+1)]%=mod;
            }   
        }
    }
    long long res=0;
    for(int i=0;i<=k;i++) {res+=dp[mm][i];res=res%mod;}
    cout<<res;
}