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

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

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

• 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.

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;}