CodeForces 632E Thief in a Shop

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codeforces & luogu

 

题目描述

A thief made his way to a shop.

As usual he has his lucky knapsack with him. The knapsack can contain k objects. There are n kinds of products in the shop and an infinite number of products of each kind. The cost of one product of kind i is ai.

The thief is greedy, so he will take exactly k products (it‘s possible for some kinds to take several products of that kind).

Find all the possible total costs of products the thief can nick into his knapsack.

 

思路

  首先只能想到这是一个完全背包,然后具体实现也不会,后来想了一会,发现可以用dp[i]来表示当总价值为i的时候,所用的物品的最小数目,由一个简单的贪心思想可以得到,显然相同的价值,用的物品越少,那么后面的发展机会就越大2333333

  但是最后发现当下面这组数据时:

  3 2

  1 2 3

  然后人就凉了,因为dp[3]跑出来是1,因为当时的dp的定义是选最小,这时就没有办法判断到底是否可以用2个物品达到价值为3。

  然后就需要百度了qwq,先把价值由小到大进行排序,然后把所有的数都减去最小的那个数,这样当dp[i]的值比k小的时候,就可以加上任意多个第一个数,这样就可以处理这种情况了,要是不看题解我估计一辈子都想不到╮(╯▽╰)╭

 

代码

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
inline int cmp(int a,int b){
    return a<b;
}
int dp[1000005];
int cos[1005];
int main(){
    int n,k;
    cin>>n>>k;
    for(register int i=0;i<=1000004;i++){
        dp[i]=99999999;
    }
    for(register int i=1;i<=n;i++){
        cin>>cos[i];
    }
    sort(cos+1,cos+n+1,cmp);
    dp[0]=0;
    for(register int i=1;i<=n;i++){
        for(register int j=cos[i]-cos[1];j<=1000004;j++){
            dp[j]=min(dp[j-(cos[i]-cos[1])]+1,dp[j]);
        }
    }
    //cout<<dp[6]<<endl;
    for(register int i=0;i<=1000004;i++){
        //cout<<dp[i]<<endl;
        if(dp[i]<=k){
            cout<<(k-dp[i])*cos[1]+dp[i]*cos[1]+i<< ;
        }
    }
}

 

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