zoukankan      html  css  js  c++  java
  • CodeForces A. Points in Segments

    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 (1lirim1≤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 lxrl≤x≤r.

    Input

    The first line of the input contains two integers nn and mm (1n,m1001≤n,m≤100) — the number of segments and the upper bound for coordinates.

    The next nn lines contain two integers each lili and riri (1lirim1≤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 5
    2 2
    1 2
    5 5
    output
    Copy
    2
    3 4
    input
    Copy
    1 7
    1 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
    ", ans);
        for(int i = 1; i <= ans; i ++) {
            printf("%d", num[i]);
            printf("%s", i != ans ? " " : "
    ");
        }
    
        return 0;
    }
    

      

  • 相关阅读:
    linux上传文件到oss的方法
    centos6.5重装python
    nfs共享文件夹
    mysql报错ERROR 2002 (HY000): Can't connect to local MySQL server through socket '/tmp/mysql.sock' (2)
    搭建网关服务器
    面试总结
    innerText兼容性问题
    Title Case
    Character frequency
    Least Common Multiple
  • 原文地址:https://www.cnblogs.com/zlrrrr/p/9845795.html
Copyright © 2011-2022 走看看