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.

竟然超时：

#include<iostream>
#include<cstdio>

using namespace std;

int f[50009],v[5009],n,m;

int main()
{
	scanf("%d%d",&n,&m);
    for(int i=1;i<=m;i++)
		scanf("%d",&v[i]);
    for(int i=1;i<=m;i++)
        for(int j=n;j>=v[i];j--)
            if(f[j]<f[j-v[i]]+v[i])
				f[j]=f[j-v[i]]+v[i];

    printf("%d",f[n]);
    return 0;
}

#include<iostream>

using namespace std;
int c,h;//c容量 h种情况
int f[50005];
int v[50005];

int main()
{
    cin >> c >> h;
    for(int i = 1;i <= h;i++)
        cin >> v[i];
    for(int i = 1;i <= h;i++)
	{
        for(int a = c;a >= v[i];a--)
		{
            if(f[a] == a)
                continue;  //此时f[a]已经取到最大值 就不用再对f[a]进行更新
            if(f[a - v[i]] + v[i] > f[a])
                f[a] = f[a - v[i]] + v[i];
        }
        if(f[c] == c)
		{//判断是否已经能够装满c体积的干草
            cout << c;//能够装满
            return 0;//退出
        }
    }
    cout << f[c];
    return 0;
}