NYOJ-【ABS】

ABS

时间限制：1000 ms  |  内存限制：65535 KB

描述

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.

输入

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.

输出

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

样例输入

3 4 1  2  2  91 -1 10 1  3  0  0  0  1  2  7  3  7  

样例输出

 8  1  6

还是自己做的题目少，这明明时裸裸的dp嘛；其实先找出来最大值，然后剩下n-1个数
分成2组，求着2组相减绝对值的最小值，别忘了把负数先找出来 

#include<cstdio>#include<algorithm>using namespace std;int v[500];int dp[500000];int abs2(int a){if(a<0)return -a;return a;}int max(int a,int b){if(a>b)return a;return b;}int main(){int t;scanf("%d",&t);while(t--){int n,i=0,a,sum=0,ans=0;scanf("%d",&n);while(n--){scanf("%d",&a);if(a>0){v[i]=a;sum+=v[i];++i;}elseans-=a;}if(i>0){sort(v,v+i);//其实也不用排序，只要找到最大值就行了 sum-=v[i-1];for(int j=0;j<=i-2;++j){for(int k=sum/2;k>=v[j];--k)dp[k]=max(dp[k],dp[k-v[j]]+v[j]);}sum=sum-2*dp[sum/2];printf("%d\n",abs2(v[i-1]+ans-abs2(sum)));}else{printf("%d\n",ans);}}return 0;}