【nyoj 1291 ABS 】

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 9
1
-1
10
1 3 0 0 0 1 2 7 3 7
样例输出
8
1
6

#include<cstdio>#include<cmath>#include<algorithm>using namespace std;int pa[210];int main(){    int T,N,i,j,a,ans,sum;    scanf("%d",&T);    while(T--)    {        scanf("%d",&N);        i=0;ans=0;sum=0;        while(N--)        {            scanf("%d",&a);            if(a<0)                ans-=a;            else                pa[++i]=a;        }        sort(pa+1,pa+1+i);        ans+=pa[i];        i--;        for(j=1;j<=i;)        {           sum=abs(pa[i]-sum);           i--;           if(j>i)            break;           sum=abs(pa[j]-sum);           j++;        }        ans-=sum;        printf("%d\n",ans);    }    return 0;}