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.

01背包。常数看着挺大但是不会T

 /*by SilverN*/

 #include<algorithm>

 #include<iostream>

 #include<cstring>

 #include<cstdio>

 #include<cmath>

 #include<vector>

 using namespace std;

 int read(){

     int x=,f=;char ch=getchar();

     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}

     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}

     return x*f;

 }

 int f[];

 int n,c,v[];

 int main(){

     c=read();n=read();

     int i,j;

     for(i=;i<=n;i++)v[i]=read();

     f[]=;

     for(i=;i<=n;i++){

         for(j=c;j>=v[i];j--){

             f[j]|=f[j-v[i]];

         }

     }

     for(j=c;j>=;j--){

         if(f[j]){

             printf("%d\n",j);

             break;

         }

     }

     return ;

 }