cable cable cable

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 152    Accepted Submission(s): 131


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 2 20 15
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

#include<stdio.h>
#include<iostream>
#include<string.h>
#include<string>
#include<ctype.h>
#include<math.h>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<bitset>
#include<algorithm>
#include<time.h>
using namespace std;
void fre() { freopen("c://test//input.in", "r", stdin); freopen("c://test//output.out", "w", stdout); }
#define MS(x, y) memset(x, y, sizeof(x))
#define ls o<<1
#define rs o<<1|1
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b > a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b < a)a = b; }
const int N = 0, M = 0, Z = 1e9 + 7, inf = 0x3f3f3f3f;
template <class T1, class T2>inline void gadd(T1 &a, T2 b) { a = (a + b) % Z; }
int casenum, casei;
int main()
{
	UL M, K;
	while (~scanf("%llu%llu", &M, &K))
	{
		UL ans = K * (M - K + 1);
		printf("%llu\n", ans);
	}
	return 0;
}
/*
【trick&&吐槽】
敢猜敢過。~沒加，收穫SB的 OLE

【題意】
有M個格子，有K個物品。我們希望在格子與物品之間連數量儘可能少的邊，使得——不論是選出M個格子中的哪K個，都可以與K個物品恰好一一匹配。

【分析】
看樣例可以猜答案，答案就是（M - K + 1） * K

從這個答案去猜測原因——
每個物品，都要向（M - K + 1）個格子連去一條邊，我們會丟棄M - K個格子，但總會剩下一個格子是與這個物品連邊的。

但是還要完成一一匹配的關係。
這個要怎麼保證呢？如果讓我們構造並輸出一組方案怎麼辦？

連邊方式就是：
1 -> [1, M - K + 1]
2 -> [2, 1 + M - K + 1]
3 -> [3, 2 + M - K + 1]
K -> [K, M]

這樣不論選哪些格子，第一個物品總是能匹配第一個格子，第二個物品總是能匹配第二個格子…… 所以總能合法匹配！

【時間複雜度&&優化】
O（1）

*/