bzo1606: [Usaco2008 Dec]Hay For Sale 购买干草

Posted 2020-10-05 a799091501

1606: [Usaco2008 Dec]Hay For Sale 购买干草

Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 1338  Solved: 991
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Description

    约翰遭受了重大的损失：蟑螂吃掉了他所有的干草，留下一群饥饿的牛．他乘着容量为C(1≤C≤50000)个单位的马车，去顿因家买一些干草．  顿因有H(1≤H≤5000)包干草，每一包都有它的体积Vi(l≤Vi≤C).约翰只能整包购买，

他最多可以运回多少体积的干草呢？

Input

    第1行输入C和H，之后H行一行输入一个Vi．

Output

 

    最多的可买干草体积．

Sample Input

7 3 //总体积为7,用3个物品来背包
2
6
5


The wagon holds 7 volumetric units; three bales are offered for sale with
volumes of 2, 6, and 5 units, respectively.

Sample Output

7 //最大可以背出来的体积

HINT

Buying the two smaller bales fills the wagon.

Source

Silver

 

题解

超级裸的01背包dp 打板子就过了（

 

/**************************************************************

    Problem: 1606

    User: a799091501

    Language: C++

    Result: Accepted

    Time:292 ms

    Memory:1608 kb

****************************************************************/

 

#include<iostream>

#include<cstdio>

#include<cstring>

#include<algorithm>

#include<cmath>

#include<queue>

#include<stack>

#define N 100001

using namespace std;

inline int read()

{

    int f=1,x=0;char ch=getchar();

    while(ch>‘9‘|ch<‘0‘)

    {

        if(ch==‘-‘)

        f=-1;

        ch=getchar();

    }

    while(ch<=‘9‘&&ch>=‘0‘)

    {

        x=(x<<3)+(x<<1)+ch-‘0‘;

        ch=getchar();

    }

    return f*x;

}

int main()

{

    int c=read(),h=read(),j,i,b[100001],v[100001];

    b[0]=1;

    for(i=1;i<=h;i++)

    {

    v[i]=read();

    for(j=c-v[i];j>=0;j--)

    if(b[j]) b[j+v[i]]=1;

    }

    int ans;

    for(i=0;i<=c;i++)

    if(b[i])ans=i;

    cout<<ans;

}