这个题目水成这个样子，，，
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 10;
int num[maxn], n;
int main()
{
scanf("%d", &n);
int sum = 0;
for (int i = 0; i < n; i++)scanf("%d", &num[i]),sum += num[i];
sort(num, num + n);
int front = 0;
for (int i = 0; i < n/2; i++)
{
front += num[i];
}
printf("%d %d",n%2==0?0:1, abs(sum - front - front));
return 0;
}
算法笔记上有这个题目的解，作者用的是quick sort找出第n/2大的数字，然后分成两堆，但是这种有一个点会超时，我每次选的都是left
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1e5 + 10;
int num[maxn], n;
int partition(int a[], int left, int right)
{
int temp = a[left];
while (left < right)
{
while (right > left&&a[right] > temp)right--;
a[left] = a[right];
while (left < right&&a[left] <= temp)left++;
a[right] = a[left];
}
a[left] = temp;
return left;
}
int main()
{
scanf("%d", &n);
int sum = 0;
for (int i = 0; i < n; i++)scanf("%d", &num[i]),sum += num[i];
int k = n / 2, left = 0, right = n - 1, mid;
while (1)
{
int pos = partition(num, left, right);
if (pos < k)left = pos + 1;
else if (pos > k) right = pos - 1;
else if (pos == k)
{
mid = pos;
break;
}
}
int front = 0;
for (int i = 0; i < mid; i++)front += num[i];
int back = sum - front;
printf("%d %d", n % 2 == 0 ? 0 : 1, back - front);
return 0;
}
