题目描述

Farmer John suffered a terrible loss when giant Australian cockroaches ate the entirety of his hay inventory, leaving him with nothing to feed the cows. He hitched up his wagon with capacity C (1 <= C <= 50,000) cubic units and sauntered over to Farmer Don's to get some hay before the cows miss a meal.

Farmer Don had a wide variety of H (1 <= H <= 5,000) hay bales for sale, each with its own volume (1 <= V_i <= C). Bales of hay, you know, are somewhat flexible and can be jammed into the oddest of spaces in a wagon.

FJ carefully evaluates the volumes so that he can figure out the largest amount of hay he can purchase for his cows.

Given the volume constraint and a list of bales to buy, what is the greatest volume of hay FJ can purchase? He can't purchase partial bales, of course. Each input line (after the first) lists a single bale FJ can buy.

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？

输入输出格式

输入格式：

• Line 1: Two space-separated integers: C and H

• Lines 2..H+1: Each line describes the volume of a single bale: V_i

输出格式：

• Line 1: A single integer which is the greatest volume of hay FJ can purchase given the list of bales for sale and constraints.

输入输出样例

7 3

2

6

5

7

说明

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

Buying the two smaller bales fills the wagon.

思路：

　　最后一个点诶。。。

　　死活过不去；；

来，上代码：

#include <cstdio>

#include <iostream>

using namespace std;

int if_z,n,m,dp[];

char Cget;

inline void in(int &now)

{

    now=,if_z=,Cget=getchar();

    while(Cget>''||Cget<'')

    {

        if(Cget=='-') if_z=-;

        Cget=getchar();

    }

    while(Cget>=''&&Cget<='')

    {

        now=now*+Cget-'';

        Cget=getchar();

    }

    now*=if_z;

}

int main()

{

    in(m),in(n);int pos;

    while(n--)

    {

        in(pos);

        //for(int i=m;i>=pos;i--) dp[i]=max(dp[i],dp[i-pos]+pos);

        for(int i=m;i>=pos;i--)

        {

            if(dp[i-pos]+pos>dp[i]) dp[i]=dp[i-pos]+pos;

        }

    }

    cout<<dp[m];

    return ;

}

正解！

#include <iostream>

using namespace std;

int n;

int main()

{

    cin>>n;

    cout<<n;

    return ;

}