构造一组解
思路:i-a[i] != j - a[j] => i - j != a[i] - a[j]这对于逆序对来说一定成立,因此从大到小排序即可
#include <iostream>
#include <algorithm>
#include <cstdio>
using namespace std;
const int N = 105;
int a[N];
int main() {
int n;
cin >> n;
while (n--) {
int m;
cin >> m;
for (int i = 0; i < m; i++) cin >> a[i];
sort(a, a + m, greater<int>());
for (int i = 0; i < m; i++) {
i && printf(" ");
printf("%d", a[i]);
}
cout << endl;
}
return 0;
}