http://codeforces.com/contest/1015/problem/A
You are given a set of nn segments on the axis OxOx, each segment has integer endpoints between 11 and mm inclusive. Segments may intersect, overlap or even coincide with each other. Each segment is characterized by two integers lili and riri (1≤li≤ri≤m1≤li≤ri≤m) — coordinates of the left and of the right endpoints.
Consider all integer points between 11 and mm inclusive. Your task is to print all such points that don‘t belong to any segment. The point xxbelongs to the segment [l;r][l;r] if and only if l≤x≤rl≤x≤r.
Input
The first line of the input contains two integers nn and mm (1≤n,m≤1001≤n,m≤100) — the number of segments and the upper bound for coordinates.
The next nn lines contain two integers each lili and riri (1≤li≤ri≤m1≤li≤ri≤m) — the endpoints of the ii-th segment. Segments may intersect, overlap or even coincide with each other. Note, it is possible that li=rili=ri, i.e. a segment can degenerate to a point.
Output
In the first line print one integer kk — the number of points that don‘t belong to any segment.
In the second line print exactly kk integers in any order — the points that don‘t belong to any segment. All points you print should be distinct.
If there are no such points at all, print a single integer 00 in the first line and either leave the second line empty or do not print it at all.
Examples
input
Copy
3 52 21 25 5
output
Copy
23 4
input
Copy
1 71 7
output
Copy
0
代码：
#include <bits/stdc++.h>
using namespace std;
int N, M;
int vis[110], num[110];
int main() {
memset(vis, 0, sizeof(vis));
scanf("%d%d", &N, &M);
for(int i = 1; i <= N; i ++) {
int l, r;
scanf("%d%d", &l, &r);
for(int j = l; j <= r; j ++)
vis[j] = 1;
}
int ans = 0;
for(int i = 1; i <= M; i ++) {
if(!vis[i]) {
ans ++;
num[ans] = i;
}
}
printf("%d\n", ans);
for(int i = 1; i <= ans; i ++) {
printf("%d", num[i]);
printf("%s", i != ans ? " " : "\n");
}
return 0;
}
原文地址：https://www.cnblogs.com/zlrrrr/p/9845795.html