A - A Compatible Pair-biaobiao88
2019-10-29 16:01:19来源:博客园 阅读 ()
A - A Compatible Pair-biaobiao88
A - A Compatible Pair
Nian is a monster which lives deep in the oceans. Once a year, it shows up on the land, devouring livestock and even people. In order to keep the monster away, people fill their villages with red colour, light, and cracking noise, all of which frighten the monster out of coming.
Little Tommy has n lanterns and Big Banban has m lanterns. Tommy's lanterns have brightness a1, a2, ..., an, and Banban's have brightness b1, b2, ..., bm respectively.
Tommy intends to hide one of his lanterns, then Banban picks one of Tommy's non-hidden lanterns and one of his own lanterns to form a pair. The pair's brightness will be the product of the brightness of two lanterns.
Tommy wants to make the product as small as possible, while Banban tries to make it as large as possible.
You are asked to find the brightness of the chosen pair if both of them choose optimally.
Input
The first line contains two space-separated integers n and m (2 ≤ n, m ≤ 50).
The second line contains n space-separated integers a1, a2, ..., an.
The third line contains m space-separated integers b1, b2, ..., bm.
All the integers range from - 109 to 109.
Output
Print a single integer — the brightness of the chosen pair.
Examples
Input2 2Output
20 18
2 14
252Input
5 3Output
-1 0 1 2 3
-1 0 1
2
Note
In the first example, Tommy will hide 20 and Banban will choose 18 from Tommy and 14 from himself.
In the second example, Tommy will hide 3 and Banban will choose 2 from Tommy and 1 from himself.
#include<iostream> #include<cmath> #include<algorithm> using namespace std; long long cmp(long long x,long long y) { return x > y; } int main() { long long a[101],b[51],c[5201],n,m; cin >> n >> m; if(n >= 2 && m <= 50) { for(int i = 0;i < n;i++) cin >> a[i]; for(int i = 0;i < m;i++) cin >> b[i]; sort(a,a + n,cmp); sort(b,b + m,cmp); for(int i = 0;i < n;i++) { c[i] = a[i] * b[0]; for(int j = 1;j < m;j++) { c[i] = max(c[i],a[i] * b[j]); } } sort(c,c + n,cmp); cout << c[1] << endl; // for(int i = 0;i < n;i++) // cout << a[i] << " "; // cout << endl; // for(int i = 0;i < m;i++) // cout << b[i] << " "; } return 0; }
排除出一个最大值的就ok了
原文链接:https://www.cnblogs.com/biaobiao88/p/11761496.html
如有疑问请与原作者联系
标签:
版权申明:本站文章部分自网络,如有侵权,请联系:west999com@outlook.com
特别注意:本站所有转载文章言论不代表本站观点,本站所提供的摄影照片,插画,设计作品,如需使用,请与原作者联系,版权归原作者所有
- JZOI5245 Competing Souls 2020-04-19
- windows10环境下QtCreator中出现skipping incompatible xxx 2020-03-31
- computer 2020-02-15
- common subsequence 2020-02-09
- C++中的字符串输入输出,转自:https://www.cnblogs.com/zzw 2019-10-30
IDC资讯: 主机资讯 注册资讯 托管资讯 vps资讯 网站建设
网站运营: 建站经验 策划盈利 搜索优化 网站推广 免费资源
网络编程: Asp.Net编程 Asp编程 Php编程 Xml编程 Access Mssql Mysql 其它
服务器技术: Web服务器 Ftp服务器 Mail服务器 Dns服务器 安全防护
软件技巧: 其它软件 Word Excel Powerpoint Ghost Vista QQ空间 QQ FlashGet 迅雷
网页制作: FrontPages Dreamweaver Javascript css photoshop fireworks Flash