做法：題目已經向你保證，輸入的x,y中，y遞增，如果y相同，x遞增。醬紫其實已經幫助我們處理好了它們的優先級。
因爲y越大，x越大的優先級越高。所以我們直接用x的值進行區間的查詢。query(x)就是找<=x位置的元素有多少個。
AC代碼：
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define IO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
#define pb(x) push_back(x)
#define sz(x) (int)(x).size()
#define sc(x) scanf("%d",&x)
#define pr(x) printf("%d\n",x)
#define abs(x) ((x)<0 ? -(x) : x)
#define all(x) x.begin(),x.end()
#define mk(x,y) make_pair(x,y)
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const double PI = 4*atan(1.0);
const int maxm = 1e8+5;
const int maxn = 1e5+5;
const int INF = 0x3f3f3f3f;
inline int read()
{
char x;
int u,flag = 0;
while(x = getchar(),x<'0' || x>'9') if(x == '-') flag = 1;
u = x-'0';
while(x = getchar(),x>='0' && x<='9') u = (u<<3)+(u<<1)+x-'0';
if(flag) u = -u;
return u;
}
int n;
int c[maxn];
int ans[maxn];
const int N = 32005;//一定要更新到這
inline int lowbit(int x){return x&(-x);}
void update(int x,int val)
{
for(int i = x;i<=N;i+=lowbit(i)) c[i]+=val;
}
int query(int x)
{
int res = 0;
for(int i=x;i;i-=lowbit(i)) res+=c[i];
return res;
}
int main()
{
#ifdef LOCAL_FILE
freopen("in.txt","r",stdin);
#endif // LOCAL_FILE
n = read();
memset(c,0,sizeof(c));
for(int i=0;i<n;i++)
{
int l,r;
l = read();r = read();
l++;
ans[query(l)]++;
update(l,1);
}
for(int i=0;i<n;i++) printf("%d\n",ans[i]);
return 0;
}