题目链接：http://codeforces.com/problemset/problem/1000/C
You are given nn segments on a coordinate line; each endpoint of every segment has integer coordinates. Some segments can degenerate to points. Segments can intersect with each other, be nested in each other or even coincide.
Your task is the following: for every k∈[1..n]k∈[1..n], calculate the number of points with integer coordinates such that the number of segments that cover these points equals kk. A segment with endpoints lili and riri covers point xx if and only if li≤x≤rili≤x≤ri.
Input
The first line of the input contains one integer nn (1≤n≤2⋅1051≤n≤2⋅105) — the number of segments.
The next nn lines contain segments. The ii-th line contains a pair of integers li,rili,ri (0≤li≤ri≤10180≤li≤ri≤1018) — the endpoints of the ii-th segment.
Output
Print nn space separated integers cnt1,cnt2,…,cntncnt1,cnt2,…,cntn, where cnticnti is equal to the number of points such that the number of segments that cover these points equals to ii.
Examples
input
3
0 3
1 3
3 8
output
6 2 1
input
3
1 3
2 4
5 7
output
5 2 0
Note
The picture describing the first example:
Points with coordinates [0,4,5,6,7,8][0,4,5,6,7,8] are covered by one segment, points [1,2][1,2] are covered by two segments and point [3][3] is covered by three segments.
The picture describing the second example:
Points [1,4,5,6,7][1,4,5,6,7] are covered by one segment, points [2,3][2,3] are covered by two segments and there are no points covered by three segments.
题目大意：给出一个坐标轴，给出n个区间，对于区间内的坐标等级+1，这些区间可能重合，叠加。。
输出每个等级的坐标的个数
一开始看到这个题，感觉很容易，但是实际写的时候却很麻烦
其实我们可以将起点和终点分别储存，因为他们是没有冲突的，例如：
[1,3],[2,5],[1,4]他们的起点分别为1，1，2。终点为3，4，5
我们把他转化成[1,5],[1,4],[2,3]也可以，因此我们就可以打乱他们的顺序，然后直接整个区间的求和。
遇见起点+1，遇见终点-1，因为是闭区间，所以终点应该往后移动一个单位：
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<map>
//#include<set>
#include<deque>
#include<queue>
#include<stack>
#include<bitset>
#include<string>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define mod 1000000007
//#define max(a,b) (a)>(b)?(a):(b)
//#define min(a,b) (a)<(b)?(a):(b)
#define clean(a,b) memset(a,b,sizeof(a))// 水印
//std::ios::sync_with_stdio(false);
struct node{
ll x;
int ves;
node(ll _x=0,ll _ves=0):x(_x),ves(_ves){}
}dot[400100];
ll ans[200100];
bool cmp(node a,node b)
{
if(a.x==b.x)
return a.ves>b.ves;
return a.x<b.x;
}
int main()
{
int n;
cin>>n;
int k=0;
for(int i=0;i<n;++i)
{
ll a,b;
cin>>a>>b;
dot[k++]=node(a,1);
dot[k++]=node(b+1,-1);
}
sort(dot,dot+k,cmp);//坐标优先，起点优先
int res=0;
for(int i=0;i<k-1;++i)
{
res=res+dot[i].ves;
if(dot[i+1].x!=dot[i].x)
ans[res]=ans[res]+dot[i+1].x-dot[i].x;
}
for(int i=1;i<=n;++i)
cout<<ans[i]<<" ";
cout<<endl;
}