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

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;
}