CF1015APoints in Segments（签到）

Posted 2021-01-14 myx12345

题意：有一条上面有n个点的数轴，给定m次操作，每次覆盖(x[i],y[i])，求最后没有被覆盖过的点的数量与他们的编号

n,m<=100

思路：

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<string>
 4 #include<cmath>
 5 #include<iostream>
 6 #include<algorithm>
 7 #include<map>
 8 #include<set>
 9 #include<queue>
10 #include<vector>
11 using namespace std;
12 typedef long long ll;
13 typedef unsigned int uint;
14 typedef unsigned long long ull;
15 typedef pair<int,int> PII;
16 typedef vector<int> VI;
17 #define fi first
18 #define se second 
19 #define MP make_pair
20 
21 const int N=1100;
22 int a[N],n,m;
23 
24 int read()
25 { 
26    int v=0,f=1;
27    char c=getchar();
28    while(c<48||57<c) {if(c==‘-‘) f=-1; c=getchar();}
29    while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
30    return v*f;
31 }
32  
33 int main()
34 {
35   //freopen("1.in","r",stdin);
36   //freopen("1.out","w",stdout);
37   scanf("%d%d",&n,&m);
38   for(int i=1;i<=n;i++) 
39   {
40       int x,y;
41     scanf("%d%d",&x,&y);
42     for(int j=x;j<=y;j++) a[j]=1;
43   }
44   int ans=0;
45   for(int i=1;i<=m;i++) ans+=a[i];
46   printf("%d
",m-ans);
47   for(int i=1;i<=m;i++)
48    if(!a[i]) printf("%d ",i); 
49   return 0;
50 }