@ 2016GCPC Knapsack in a Globalized World (DP 完全背包)

最新推荐文章于 2024-08-18 14:41:12 发布

sizaif

最新推荐文章于 2024-08-18 14:41:12 发布

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分类专栏： ACM进阶之路 # 动态规划相关算法文章标签： DP 完全背包因子

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在全球化的影响下，传统的盗窃行为已经发生了变化。新的游戏规则要求窃贼必须针对大型商店行动，利用无限的商品供应，携带巨大的背包，并确保背包完全装满。本文探讨了如何通过编程来决定是否应该掠夺一家商店，提出了一种解决大背包容量问题的方法，即通过寻找背包容量的因子来判断是否能够使用给定的物品类型恰好填满背包。

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Globalization stops at nothing, not even at the good old honest profession of a burglar. Nowadays it is not enough to break in somewhere, take everything you can carry and dart off. No! You have to be competitive, optimize your proﬁt and utilize synergies.
So, the new game rules are:
• break only into huge stores, so there is practically endless supply of any kind of items;
• your knapsack should be huge;
• your knapsack should be full (there should be no empty space left).
Damn you, globalization, these rules are not easy to follow! Luckily, you can write a program, which will help you decide whether you should loot a store or not.

输入

The input consists of:
• one line with two integers n (1 ≤ n ≤ 20) and k (1 ≤ k ≤ 1018 ), where n is the number of different item types and k is the size of your knapsack;
• one line with n integers g 1 , . . . , gn (1 ≤ gi ≤ 103 for all 1 ≤ i ≤ n), where g1 , . . . , gn are the sizes of the n item types.

输出

Output “possible” if it is possible to ﬁll your knapsack with items from the store (you mayassume that there are enough items of any type), otherwise output “impossible”.

样例输入

2 10000000000
3 6

样例输出

impossible

[题意]

给 n 个物品和背包容量 ( 容量很大) 物品数量无限

问能不能用这 n 个物品 (其中一部分或全部) 恰好填满背包.,

[显然一个完全背包问题]

但是背包的容量很大, ll 级别. dp 不出来.

但是我们可以去 dp 背包的因子啊. 如果他的因子能被背出来., 那么他一定可以. 倍数关系...

认为因子在 2e6的范围内都能找到 k 的因子.

所以只需要 dp 到 2e6 就可以.

[代码]

#include <bits/stdc++.h>
#include <stdio.h>
#define rep(i,a,n) for(int i=a;i<=n;i++)
#define per(i,a,n) for(int i=n;i>=a;i--)

typedef long long ll;
const int maxn = 2e6+10;
const int mod =1e9+7;
const int inf = 0x3f3f3f3f;
using namespace std;

int c[100];
int dp[maxn];
int main(int argc, char const *argv[])
{
	ll n,k;
	scanf("%lld %lld",&n,&k);
	rep(i,1,n) scanf("%d",&c[i]);
	memset(dp,0,sizeof(dp));
	dp[0] = 1;
	for(int i = 1; i <= n; i++)
	{
    	for(int j = 0  ; j <= maxn-1 ; j ++ )
    	{
        	if( dp[j] && j + c[i] <= maxn-1 )
        		dp[j+c[i]] = 1;
		}
	}
	int flag = 0;
	rep(i,1,maxn-1)
	{
		if( dp[i] && k%i==0)
		{
			flag = 1;
			break;
		}
	}
	if( flag) printf("possible\n");
	else printf("impossible\n");

	return 0;
}