hdu6195(数学)

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

3 220 15

 

Sample Output

490
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.
题意：给你M个显示屏，K种颜色，用最少的电缆线使得其中任意K个显示屏显示不同颜色。
思路：我是这样想的，先将其中的K个用一根线连接，需要K根线，剩下的M-K个每个都必须连接K种颜色，需要(M-K)*K根线。
代码：
#include<bits/stdc++.h>using namespace std;const int maxn=1e5+5;typedef long long LL;int main(){    LL m,k;    while(scanf("%lld%lld",&m,&k)!=EOF)    {        LL num1=m-k;        LL ans=num1*k+k;        printf("%lld\n",ans);    }    return 0;}