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最新推荐文章于 2025-03-03 13:47:25 发布

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本文介绍了一个程序员团队在面对特定代码量任务时，如何通过合理的任务分配来确保代码总错误数在限定范围内。文章详细解释了一个算法流程，该流程能够帮助团队计算出所有可行的任务分配方案数量，并最终给出在给定模数条件下的方案余数。

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Programmers working on a large project have just received a task to write exactly mlines 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

注释见代码里：

#include<algorithm>
#include<iostream>
using namespace std;

/*
* 此题的每个程序员对应的bug是指这个程序员打每行必定会错这么多个，而不是最多；
*line[j][5]值得是打了j行，出现了5个bug
*/

int main()
{
	int n, m, b, mod, i, j, k;  //用k代表bug的数量
	int bug[501] , line[501][501];  //line储存种类数
	long long sum=0;
	memset(line, 0, sizeof(line));
	cin >> n >> m >> b >> mod;
	for (i = 1; i <= n; i++) {
		cin >> bug[i];
	}
	line[0][0] = 1;  //0行0个bug时方案1个
	for (i = 1; i <= n; i++) {  //i表示第几个人,i不重要，重要的是bug[i]
		for (j = 1; j <= m; j++) {
			if (bug[i] <= b) {
				for (k = bug[i]; k <= b; k++) {
					line[j][k] = line[j - 1][k - bug[i]] % mod + line[j][k] % mod;  //新的方案数加上以前的方案数，当程序员的bug相同时，交换顺序也会
					line[j][k] = line[j][k] % mod;                                  //增加方案种类
				}
			}
		}
	}
	for (i = 0; i <= b; i++) {  //m行时，不超过b个bug的方案累加
		cout << "敲了2行代码，错" << i << "个bug的方案有：";
		cout << line[2][i]<<endl;
		sum = (sum + line[m][i]) % mod;
	}
	cout << "最后的结果是："<<sum;
	return 0;
}

结果：