poj 1187 表示范围的线性dp

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原创最新推荐文章于 2026-05-10 09:11:42 发布 234 阅读 AI 写同款· GEO 优化 ›

从一次仿真冲突调试说起：手把手分析Verilog里wire信号被‘意外’拉成x值的完整流程本文深入探讨了Verilog中wire信号意外变为x值的调试流程，重点分析了多驱动冲突和强度建模问题。通过实际案例和系统化的调试方法，帮助工程师快速定位和解决信号冲突，提升数字电路设计的可靠性。阅读详情

#include <cstdio>
#include <iostream>
#include <iomanip>
#include <string>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <set>
#include <vector>
#include <map>
#include <algorithm>
#include <cmath>
#include <stack>

#define INF 0x3f3f3f3f
#define IMAX 2147483646
#define LINF 0x3f3f3f3f3f3f3f3f
#define ll long long
#define ull unsigned long long
#define uint unsigned int

using namespace::std;

const int mod = 11380;
int f[11][11][11][31];

int dfs(int l1, int l2, int l3, int deep) {
	if (f[l1][l2][l3][deep] > -1)return f[l1][l2][l3][deep];
	if (l1 + l2 + l3 == 0)return f[l1][l2][l3][deep] = 1;
	if (deep == 0)return f[l1][l2][l3][deep] = 0;
	int ans = 0;
	for (int i = 0; i <= l3; i++) {
		if (i)  ans = (ans + dfs(0, 0, i - 1, deep - 1) * dfs(l1, l2, l3 - i, deep)) % mod;
		for (int j = 0; j <= l2; j++) {
			if (j)  ans = (ans + dfs(0, j - 1, i, deep - 1) * dfs(l1, l2 - j, l3 - i, deep)) % mod;
			for (int k = 1; k <= l1; k++) 
				ans = (ans + dfs(k - 1, j, i, deep - 1) * dfs(l1 - k, l2 - j, l3 - i, deep)) % mod;
		}
	}
	return f[l1][l2][l3][deep] = ans;
}

int a, b,c, d;

int main() {
	memset(f, -1, sizeof(f));
	scanf("%d%d%d%d", &a, &b, &c, &d);
	dfs(a, b, c, d);
	if (d) {
		dfs(a, b, c, d - 1);
		printf("%d\n", (f[a][b][c][d] - f[a][b][c][d - 1] + mod) % mod);
		return 0;
	}
	printf("%d\n", f[a][b][c][d]);


	return 0;
}