HDU6195 cable cable cable

最新推荐文章于 2021-08-29 12:34:22 发布

Jelly_acm

最新推荐文章于 2021-08-29 12:34:22 发布

阅读量452

点赞数

分类专栏： HDU ----水题

本文链接：https://blog.csdn.net/Jelly_acm/article/details/77925530

版权

HDU 同时被 2 个专栏收录

181 篇文章

订阅专栏

----水题

75 篇文章

订阅专栏

cable cable cable

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 90 Accepted Submission(s): 79

Problem Description

Connecting the display screen and signal sources which produce different color signals by cables, then the display screen can show the color of the signal source.Notice that every signal source can only send signals to one display screen each time.
Now you have

M display screens and

K different signal sources(

K≤M≤232−1 ). Select

K display screens from

M display screens, how many cables are needed at least so that **any**

K display screens you select can show exactly

K different colors.

Input

Multiple cases (no more than

100 ), for each test case:
there is one line contains two integers

M and

K .

Output

Output the minimum number of cables

N .

Sample Input

Sample Output

      
      
       
       4
90

       
       
        
        
         
         Hint
        
        

        
        
         
         
        
        
As the picture is shown, when you select M1 and M2, M1 show the color of K1, and M2 show the color of K2.

When you select M3 and M2, M2 show the color of K1 and M3 show the color of K2.

When you select M1 and M3, M1 show the color of K1.

Source

2017 ACM/ICPC Asia Regional Shenyang Online

——————————————————————————————————

题目的意思是给出m台显示器和k种颜色，要求每台显示器匹配一或多种颜色，

使得任意的选出k台都能颜色不相同

思路：找规律

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF = 0x3f3f3f3f;

int main()
{
    LL k,m;
    while(~scanf("%lld%lld",&m,&k))
    {
        LL temp=m-k+1;
        printf("%lld\n",temp*k);
    }
    return 0;
}