题意: 找到一个数列a的每个i位置前面比a[i]小的元素个数。
思维: 数据比较小(n最多才100),可直接暴力遍历判断下就可。
代码实现:
#include<bits/stdc++.h>
#define endl '\n'
#define null NULL
#define ll long long
#define int long long
#define pii pair<int, int>
#define lowbit(x) (x &(-x))
#define ls(x) x<<1
#define rs(x) (x<<1+1)
#define me(ar) memset(ar, 0, sizeof ar)
#define mem(ar,num) memset(ar, num, sizeof ar)
#define rp(i, n) for(int i = 0, i < n; i ++)
#define rep(i, a, n) for(int i = a; i <= n; i ++)
#define pre(i, n, a) for(int i = n; i >= a; i --)
#define IOS ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
const int way[4][2] = {{1, 0}, {-1, 0}, {0, 1}, {0, -1}};
using namespace std;
const int inf = 0x7fffffff;
const double PI = acos(-1.0);
const double eps = 1e-6;
const ll mod = 1e9 + 7;
const int N = 2e5 + 5;
int n, a[200];
signed main()
{
IOS;
cin >> n;
for(int i = 1; i <= n; i ++) cin >> a[i];
for(int i = 1; i <= n; i ++){
int x = 0;
for(int j = 1; j < i; j ++)
if(a[j] < a[i]) x ++;
cout << x << (i == n ? '\n' : ' ');
}
return 0;
}