HDU 6159 cable cable cable （规律水题）

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博客围绕显示屏与信号源连线问题展开。已知有M个显示屏和K个信号源，需找出最少的电缆数，使任意选取的K个显示屏能显示K种不同颜色。思路是先让K个显示屏各连一个信号源，剩余（M - K）个显示屏每个都连K个信号源。

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=6195

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 MM display screens and KK different signal sources(K≤M≤232−1K≤M≤232−1). Select KK display screens from MM display screens, how many cables are needed at least so that **any** KK display screens you select can show exactly KK different colors.

Input

Multiple cases (no more than 100100), for each test case:
there is one line contains two integers MM and KK.

Output

Output the minimum number of cables NN.

Sample Input
3 2
20 15
Sample Output
4
90

        
  

题意：

每个信号源都有不同的颜色，有M个显示屏和K个信号源，求最少的连线使得k个显示屏有k种颜色

思路：

先取k个显示屏各连一个信号源。剩下的（m-k）个显示屏必然每个都连k个信号源才能满足条件。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
int main()
{
    ll n,k;
    while(~scanf("%lld%lld",&n,&k)){
        printf("%lld\n",(n-k+1)*k);
    }
    return 0;
}