思路
先把数组从小到大排序,然后把前n / 2个元素累加到s1上,把后边的元素累加到s2上;
不需要考虑奇偶的问题,因为无论奇数还是偶数,s1总是前n/2个元素的总和;
C++ 代码
#include <iostream>
#include <algorithm>
#include <cmath>
using namespace std;
const int N = 1e5 + 10;
int a[N];
int main(){
int n, tmp = 0, mid = 0, s1 = 0, s2 = 0;
cin >> n;
int flag = n % 2;
for(int i = 0; i < n; i++) scanf("%d", &a[i]);//读入数据
sort(a, a + n);//从小到大排序
//依次累加
for(int i = 0; i < n; i++) {
if(i < n / 2) s1 += a[i];
else s2 += a[i];
}
printf("%d %d\n", flag, s2 - s1);
return 0;
}