describe
Mr.Ha is a famous scientist .He has just not got a kind of magic
medicine called Entropy Cracker.The medicine was preserved in V
bottles,and the i-th (1≤i≤N) bottle contains V liters of medicine.To
make the magic available, he needs to put the medicine from all the
bottles together.Due to mixing medicine is a dangerous action, Mr.Ha takes a huge
container with infinite volume and decides only to pour a whole bottle
of medicine into the container each time.After adding a p liter`s
bottle of medicine into the container with q liters of medicinein
it,the resulted volume of medicine in the container will be |p-q|
liters.Initially the container is empty ,and Mr.Ha can put the bottles of
medicine into the container by arbitrary order.Finally if there are R
liters of medicine in the container ,Mr.Ha will be able to use the
magic to increase the time for R seconds every day so that he can
achieve more work! Help Mr.Ha to make an arrangement so that the
resulted R is maximum.
enter
The first line contains an integer T,indicating the number of test
cases. For each test case ,the first line contains an integer N
1≤N≤200,indicating the number of botters, and the second line contains
V (|vi|<=500),indicating the volume of medicine in each bottle
Attention the volume may be negative as a result of magic The sum of
N in all test cases will not exceed 2000.
output
For each test case , output the case number in the format of the
format of the sample ,and an Integer ,the maximum seconds Mr.Ha will
able to increase
sample input
3
4
1 2 2 9
1
-1
10
1 3 0 0 0 1 2 7 3 7
Sample output
8
1
6
ideas
Let’s talk about the meaning of the question first. There is a container with infinite capacity. I will give you na potion. Each potion has a weight (which can be negative). To add a potion to the container, after each addition, the weight in the container will be It becomes the absolute value of the difference between the weight of the container and the weight of the ppotion .q
You can add potions in any order and ask what is the maximum weight of potions in the final container.
We consider that negative numbers must be added last, so we directly store the negative numbers and subtract them first (subtracting negative numbers is equal to adding positive numbers).
Put all the positive numbers in an array and sort them from large to small. Here we do a greedy one, the largest one can definitely be put last, this situation must be optimal, we just need to find a way to make them The smallest is enough. We can consider dividing it into two piles to minimize the difference between the two piles, so that the problem is transformed into nyoj邮票分你一半this problem, only need to carry out the 01 backpack once, and then sum-dp[v]the difference between the two piles.
code
#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <sstream>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#include<list>
using namespace std;
#define mem(a,b) memset(a,b,sizeof(a))
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define inf 0x3f3f3f3f
typedef long long ll;
const int N=50000+20;
int a[N],dp[N];
bool cmp(int x,int y)
{
return x>y;
}
int main()
{
int t,n,x;
scanf("%d",&t);
while(t--)
{
mem(a,0);
mem(dp,0);
scanf("%d",&n);
int ans=0,len=0,sum=0;
for(int i=1; i<=n; i++)
{
scanf("%d",&x);
if(x<0)ans-=x;
else
{
a[len++]=x;
sum+=x;
}
}
sort(a,a+len,cmp);
ans+=a[0];
sum-=a[0];
int v=sum/2;
for(int i=1; i<len; i++)
for(int j=v; j>=a[i]; j--)
dp[j]=max(dp[j],dp[j-a[i]]+a[i]);
sum=sum-2*dp[v];
printf("%d\n",ans-sum);
}
return 0;
}