[第三次训练]A simple problem

A simple problem

Description

Dreamone has a lovely cat. The cat can comfort you if you are very dejected, she can also play with you if 

you are very bored. So Dreamone loves her very much. Of course, you bet, she can also make you trouble if 

she wants to be naughty. This time, she comes up with a simple problem for fun, just like this: 

She has 2n toys. In her heart, she has used a number to stand for each toy, and for example, the 2n toys are 

given like this: 

                       A1、A2、A3、….A2n-1、A2n 

She wants to divide them into two groups, where each group contains n toys. So we can use B1、

B2...Bn ,which are all from A1、 A2...A2n ,to stand for Group One ,use C1、 C2….Cn ,which are all from A1、

A2...A2n, to stand for Group Two. The cat wants to know the minimum value S for the expression below: 

S=|B1-C1|+|B2-C2|+|B3-C3|+...+|Bn-Cn| 

As we know Dreamone is studying MaJiang recently, so he has no time to solve the simply problem. But he 

knows the 6th program contest of SWUST is on, so he turns to you for help. Can you help him?

Input

The first line of input will be a positive integer C indicating how many data sets will be included. Each of 

the C data sets will contain two parts：The first part contains a number n(1<=n<=100000),and the second 

parts contains 2n numbers ,which are A1、 A2、 A2n-1、 A2n (0<=Ai<=1000(1<=i<=2n)),represented 2n toys 

Output

For each case, output the minimum value S for answer.  

Sample Input

2 1 1 3 2 1 1 1 2

Sample Output

2 1

HINT

解题报告

此题题意并不难理解，就是从A2n个玩具中分成两堆，分别是B1-Bn和C1-Cn，然后套用公式算出最小的S。此题只需要用sort排序一下，然后相邻两个依次相减，即为最小的S，代码如下：

#include<iostream>#include<cstdio>#include<algorithm>#include<cmath>using namespace std;int a[200010];int main(){    int n,x,i,S;    scanf("%d",&n);    while(n--)    {        S=0;        scanf("%d",&x);        for(i=0;i<2*x;i++)            scanf("%d",&a[i]);        sort(a,a+2*x);        for(i=0;i<2*x;i+=2)            S+=fabs(a[i]-a[i+1]);        printf("%d\n",S);    }    return 0;}