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

C. Writing Code

Time Limit: 20 Sec  Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/544/problem/C

Description

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 Input

3 3 3 100
1 1 1

Sample Output

10

HINT

题意

有n个程序员，第i个程序员每行会有ai个bug，然后让你按着顺序安排程序员，问你有多少种安排方法，可以使得写m行的bug最多b个

题解：

简单dp，dp[i][j]表示写i行bug数有j个的方法数

转移方程和背包问题类似

最后扫一遍统计一下就好

代码：

//qscqesze
#include <cstdio>
#include <cmath>
#include <cstring>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <set>
#include <vector>
#include <sstream>
#include <queue>
#include <typeinfo>
#include <fstream>
#include <map>
#include <stack>
typedef long long ll;
using namespace std;
//freopen("D.in","r",stdin);
//freopen("D.out","w",stdout);
#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)
#define maxn 1000001
#define eps 1e-9
int Num;
char CH[];
//const int inf=0x7fffffff;   //нчоч╢С
const int inf=0x3f3f3f3f;
/*

inline void P(int x)
{
    Num=0;if(!x){putchar('0');puts("");return;}
    while(x>0)CH[++Num]=x%10,x/=10;
    while(Num)putchar(CH[Num--]+48);
    puts("");
}
*/
inline ll read()
{
    int x=,f=;char ch=getchar();
    while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
    while(ch>=''&&ch<=''){x=x*+ch-'';ch=getchar();}
    return x*f;
}
inline void P(int x)
{
    Num=;if(!x){putchar('');puts("");return;}
    while(x>)CH[++Num]=x%,x/=;
    while(Num)putchar(CH[Num--]+);
    puts("");
}
//**************************************************************************************

int dp[][];
int main()
{
    int n=read(),m=read(),b=read(),mod=read();
    dp[][]=;
    for(int i=;i<=n;i++)
    {
        int x=read();
        for(int j=;j<=m;j++)
            for(int t=x;t<=b;t++)
                dp[j][t]=(dp[j][t]+dp[j-][t-x])%mod;
    }
    int ans=;
    for(int i=;i<=b;i++)
        ans=(ans+dp[m][i])%mod;
    cout<<ans<<endl;
}

C. Writing Code