Codeforce-A-Two distinct points(暴力)
output
standard output
You are given two segments [l1;r1][l1;r1] and [l2;r2][l2;r2] on the xx-axis. It is guaranteed that l1<r1l1<r1 and l2<r2l2<r2. Segments may intersect, overlap or even coincide with each other.
The example of two segments on the xx-axis.
Your problem is to find two integers aa and bb such that l1≤a≤r1l1≤a≤r1, l2≤b≤r2l2≤b≤r2 and a≠ba≠b. In other words, you have to choose two distinct integer points in such a way that the first point belongs to the segment [l1;r1][l1;r1] and the second one belongs to the segment [l2;r2][l2;r2].
It is guaranteed that the answer exists. If there are multiple answers, you can print any of them.
You have to answer qq independent queries.
Input
The first line of the input contains one integer qq (1≤q≤5001≤q≤500) — the number of queries.
Each of the next qq lines contains four integers l1i,r1i,l2il1i,r1i,l2i and r2ir2i (1≤l1i,r1i,l2i,r2i≤109,l1i<r1i,l2i<r2i1≤l1i,r1i,l2i,r2i≤109,l1i<r1i,l2i<r2i) — the ends of the segments in the ii-th query.
Output
Print 2q2q integers. For the ii-th query print two integers aiai and bibi — such numbers that l1i≤ai≤r1il1i≤ai≤r1i, l2i≤bi≤r2il2i≤bi≤r2i and ai≠biai≠bi. Queries are numbered in order of the input.
It is guaranteed that the answer exists. If there are multiple answers, you can print any.
Example
input
Copy
5
1 2 1 2
2 6 3 4
2 4 1 3
1 2 1 3
1 4 5 8
output
Copy
2 1
3 4
3 2
1 2
3 7
代码:
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int main()
{
int n;
cin>>n;
int l1,r1,l2,r2;
for(int t=0;t<n;t++)
{
scanf("%d%d%d%d",&l1,&r1,&l2,&r2);
int fi=l1;
int se;
for(int t=l2;t<=r2;t++)
{
if(t!=fi)
{
se=t;
break;
}
}
cout<<fi<<" "<<se<<endl;
}
return 0;
}
Codeforce-A-Two distinct points(暴力)的更多相关文章
- Two distinct points CodeForces - 1108A (签到)
You are given two segments [l1;r1][l1;r1] and [l2;r2][l2;r2] on the xx-axis. It is guaranteed that l ...
- [CodeForces-1036E] Covered Points 暴力 GCD 求交点
题意: 在二维平面上给出n条不共线的线段,问这些线段总共覆盖到了多少个整数点 解法: 用GCD可求得一条线段覆盖了多少整数点,然后暴力枚举线段,求交点,对于相应的 整数交点,结果-1即可 #inclu ...
- Codeforces 850A - Five Dimensional Points(暴力)
原题链接:http://codeforces.com/problemset/problem/850/A 题意:有n个五维空间内的点,如果其中三个点A,B,C,向量AB,AC的夹角不大于90°,则点A是 ...
- Codeforce Gym 100015I Identity Checker 暴力
Identity Checker 题目连接: http://codeforces.com/gym/100015/attachments Description You likely have seen ...
- codeforce 985B Switches and Lamps(暴力+思维)
Switches and Lamps time limit per test 3 seconds memory limit per test 256 megabytes input standard ...
- CF1108A Two distinct points 题解
Content 有 \(q\) 次询问,每次询问给定四个数 \(l_1,r_1,l_2,r_2\).对于每次询问,找到两个数 \(a,b\),使得 \(l_1\leqslant a\leqslant ...
- Codeforces Round #394 (Div. 2) B. Dasha and friends 暴力
B. Dasha and friends 题目连接: http://codeforces.com/contest/761/problem/B Description Running with barr ...
- Codeforces Round #394 (Div. 2)A水 B暴力 C暴力 D二分 E dfs
A. Dasha and Stairs time limit per test 2 seconds memory limit per test 256 megabytes input standard ...
- Codeforces Round #394 (Div. 2) B. Dasha and friends —— 暴力 or 最小表示法
题目链接:http://codeforces.com/contest/761/problem/B B. Dasha and friends time limit per test 2 seconds ...
随机推荐
- 计算数组arr中所有元素的和
<!DOCTYPE html> <html> <head> <meta charset="utf-8" /> <title&g ...
- python爬虫(6)--Requests库的用法
1.安装 利用pip来安装reques库,进入pip的下载位置,打开cmd,默认地址为 C:\Python27\Scripts 可以看到文件中有pip.exe,直接在上面输入cmd回车,进入命令行界面 ...
- python爬虫--编码问题y
1)中文网站爬取下来的内容中文显示乱码 Python中文乱码是由于Python在解析网页时默认用Unicode去解析,而大多数网站是utf-8格式的,并且解析出来之后,python竟然再以Unicod ...
- android手机分辨率的一些说明
Android上常见度量单位 px(像素):屏幕上的点,绝对长度,与硬件相关 in(英寸):长度单位 mm(毫米):长度单位 pt(磅):1/72英寸,point dp(与密度无关的像素):一种基于屏 ...
- 线段树教做人系列(2)HDU 4867 XOR
题意:给你一个数组a,长度为.有两种操作.一种是改变数组的某个元素的值,一种是满足某种条件的数组b有多少种.条件是:b[i] <= a[i],并且b[1]^b[2]...^b[n] = k的数组 ...
- [转] php foreach用法和实例
PHP 4 引入了 foreach 结构,和 Perl 以及其他语言很像.这只是一种遍历数组简便方法.foreach 仅能用于数组,当试图将其用于其它数据类型或者一个未初始化的变量时会产生错误.有两种 ...
- 面试题: 1天的java面试题 已看1
1,自我介绍下,我直接说的项目经历,(哪年在哪个公司呆过) 2,问是否有带过团队的经历,我说去年带过一次. 3,Struts是单例模式还是多例模式?我先说单例模式,后说多例模式. Struts1是单例 ...
- HDOJ 4802 GPA
Problem Description In college, a student may take several courses. for each course i, he earns a ce ...
- 回溯法和DFS leetcode Combination Sum
代码: 个人浅薄的认为DFS就是回溯法中的一种,一般想到用DFS我们脑中一般都有一颗解法树,然后去按照深度优先搜索去寻找解.而分支界限法则不算是回溯,无论其是采用队列形式的还是优先队列形式的分支界限法 ...
- 【转】 robotframework(rf)中对时间操作的datetime库常用关键字
转自http://blog.csdn.net/r455678/article/details/52993765 DateTime库是robotframework内置的库 1.对固定日期进行操作,增加或 ...