AC日记——[USACO08DEC]干草出售Hay For Sale 洛谷 P2925

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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

输入样例#1:
7 3 
2 
6 
5 
输出样例#1:
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[50005];

char Cget;

inline void in(int &now)
{
    now=0,if_z=1,Cget=getchar();
    while(Cget>9||Cget<0)
    {
        if(Cget==-) if_z=-1;
        Cget=getchar();
    }
    while(Cget>=0&&Cget<=9)
    {
        now=now*10+Cget-0;
        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 0;
}

 

正解!

#include <iostream>

using namespace std;

int n;

int main()
{
    cin>>n;
    cout<<n;
    return 0;
}

 

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