NYOJ 1289 ABS 【贪心】

ABS

时间限制：1000 ms  |  内存限制：65535 KB

难度：3

 

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

来源

某校校赛

上传者

MQLYES

我们要求所有数的差绝对值。假设目前都为正数（全是负数等效于正数，有负数有正数就把负数加到最值里），这几个数进行完运算的值一定小于等于最大的那个值，现在，我们就要把最大的拿下来，让其他的数进行运算使其值最小，让最大值减去就是答案。
对于剩下的部分，每次取区间的两个端点做差，得到的是一个相对大的值，再用它与剩下的元素中最大的做差，再和最小的做差，减小这个值，重复上面的过程。
这是我的思路，不会证明，只是测了一些数据，不知道对不对，可能是数据比较水，可能是对了，求思路，求数据hack。题解好像说用dp解决，明天看看。
代码是比赛时候写的，比较乱。

 

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int n, m,ar[205], a, an[205];
int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &m); int num = 0;
        int cnt = 0; n = 0;
        for (int i = 0; i < m; i++) {
            scanf("%d", &a);
            if (a < 0) an[cnt++] = a;
            else ar[n++] = a;
        }
        if (!n && cnt) {
            n = cnt;
            for (int i = 0; i < n; i++) {
                ar[i] = -an[i];
            }
        }
        else if (n && cnt) {
            for (int i = 0; i < cnt; i++) {
                num -= an[i];
            }
        }
        sort(ar, ar + n); int ans = 0;
        for (int i = 0; i < (n - 1)/2; i++) {
            ans = abs(ans - ar[n - i - 2]);
            ans = abs(ans - ar[i]);
        }
        if (n%2 == 0) ans = abs(ans - ar[n/2 - 1]);
        printf("%d
", abs(ar[n - 1]+num - ans));
    }
    return 0;
}