[二进制完全多重背包] Transport Ship 计蒜客-优快云博客

本文链接：https://blog.youkuaiyun.com/ummmmm/article/details/82726941

本文探讨了一种算法，用于计算使用不同类型的运输船只，在确保每艘船满载的情况下，运输特定总重量货物的方案数量。算法通过01背包问题的优化解决，考虑了船只类型及其承载能力，同时应用了模运算处理可能的大数值结果。

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26.89%
1000ms
65536K

There are NN different kinds of transport ships on the port. The i^{th}ith kind of ship can carry the weight of V[i]V[i] and the number of the i^{th}ith kind of ship is 2^{C[i]} - 12C[i]−1. How many different schemes there are if you want to use these ships to transport cargo with a total weight of SS?

It is required that each ship must be full-filled. Two schemes are considered to be the same if they use the same kinds of ships and the same number for each kind.

Input

The first line contains an integer T(1 \le T \le 20)T(1≤T≤20), which is the number of test cases.

For each test case:

The first line contains two integers: N(1 \le N \le 20), Q(1 \le Q \le 10000)N(1≤N≤20),Q(1≤Q≤10000), representing the number of kinds of ships and the number of queries.

For the next NN lines, each line contains two integers: V[i](1 \le V[i] \le 20), C[i](1 \le C[i] \le 20)V[i](1≤V[i]≤20),C[i](1≤C[i]≤20), representing the weight the i^{th}ith kind of ship can carry, and the number of the i^{th}ith kind of ship is 2^{C[i]} - 12C[i]−1.

For the next QQ lines, each line contains a single integer: S(1 \le S \le 10000)S(1≤S≤10000), representing the queried weight.

Output

For each query, output one line containing a single integer which represents the number of schemes for arranging ships. Since the answer may be very large, output the answer modulo 10000000071000000007.

样例输入复制

样例输出复制

0
1

题目来源

ACM-ICPC 2018 焦作赛区网络预赛

#include <bits/stdc++.h>
using namespace std;

const int mod = 1000000007;

int v[25], c[25];
int a[25 * 25];
int dp[10010];

int main()
{
	int T;
	scanf ("%d", &T);
	while (T--)
	{
		int n, q;
		scanf ("%d %d", &n, &q);
		for (int i = 1; i <= n; i++)
			scanf("%d %d", &v[i], &c[i]);
		//数量: 2 ^ n - 1 == 2 ^ 0 + 2 ^ 1 + 2 ^ 2 + ... + 2 ^ (n - 1)
		
		// 二进制枚举优化
		// 将 2 ^ n - 1 艘船分成 2 ^ 0 \ 2 ^ 1 \ 2 ^ 2 \ ... \ 2 ^ (n - 1) 各个部分
		// 可组成所有情况的船数
		int num = 0;
		for (int i = 1; i <= n; i++)
		{
			for (int j = 0; j < c[i]; j++)
				a[num++] = v[i] * (1 << j);
		}
		
		// 01背包
		memset(dp, 0, sizeof dp);
		dp[0] = 1;
		for (int i = 0; i < num; i++)
		{
			for (int j = 10000; j >= a[i]; j--)
				dp[j] = (dp[j] + dp[j - a[i]]) % mod;
		}
		
		while (q--)
		{
			int que;
			scanf("%d", &que);
			printf("%d\n", dp[que] % mod);
		}
	}
	return 0;
}