From f1558bb8d312178b341b267376e5bbb996c3cfdc Mon Sep 17 00:00:00 2001
From: Santo Cariotti <sancn@live.com>
Date: Tue, 2 May 2017 09:19:13 +0200
Subject: Deleted Facebook Hacker Cup files

---
 sluggish_security.cpp | 203 --------------------------------------------------
 1 file changed, 203 deletions(-)
 delete mode 100644 sluggish_security.cpp

(limited to 'sluggish_security.cpp')

diff --git a/sluggish_security.cpp b/sluggish_security.cpp
deleted file mode 100644
index 33f30b0..0000000
--- a/sluggish_security.cpp
+++ /dev/null
@@ -1,203 +0,0 @@
-// Hacker Cup 2017
-// Round 3
-// Sluggish Security
-// Jacob Plachta
-
-#include <algorithm>
-#include <functional>
-#include <numeric>
-#include <iostream>
-#include <iomanip>
-#include <cstdio>
-#include <cmath>
-#include <complex>
-#include <cstdlib>
-#include <ctime>
-#include <cstring>
-#include <cassert>
-#include <string>
-#include <vector>
-#include <list>
-#include <map>
-#include <set>
-#include <deque>
-#include <queue>
-#include <stack>
-#include <bitset>
-#include <sstream>
-using namespace std;
-
-#define LL long long
-#define LD long double
-#define PR pair<int,int>
-
-#define Fox(i,n) for (i=0; i<n; i++)
-#define Fox1(i,n) for (i=1; i<=n; i++)
-#define FoxI(i,a,b) for (i=a; i<=b; i++)
-#define FoxR(i,n) for (i=(n)-1; i>=0; i--)
-#define FoxR1(i,n) for (i=n; i>0; i--)
-#define FoxRI(i,a,b) for (i=b; i>=a; i--)
-#define Foxen(i,s) for (i=s.begin(); i!=s.end(); i++)
-#define Min(a,b) a=min(a,b)
-#define Max(a,b) a=max(a,b)
-#define Sz(s) int((s).size())
-#define All(s) (s).begin(),(s).end()
-#define Fill(s,v) memset(s,v,sizeof(s))
-#define pb push_back
-#define mp make_pair
-#define x first
-#define y second
-
-template<typename T> T Abs(T x) { return(x<0 ? -x : x); }
-template<typename T> T Sqr(T x) { return(x*x); }
-
-const int INF = (int)1e9;
-const LD EPS = 1e-9;
-const LD PI = acos(-1.0);
-
-bool Read(int &x)
-{
-	char c,r=0,n=0;
-	x=0;
-		for(;;)
-		{
-			c=getchar();
-				if ((c<0) && (!r))
-					return(0);
-				if ((c=='-') && (!r))
-					n=1;
-				else
-				if ((c>='0') && (c<='9'))
-					x=x*10+c-'0',r=1;
-				else
-				if (r)
-					break;
-		}
-		if (n)
-			x=-x;
-	return(1);
-}
-
-#define LIM 2000005
-#define MOD 1000000007
-
-int A[LIM],B[LIM],S[LIM];
-LL fct[LIM],ifct[LIM];
-
-PR gcd(int a,int b)
-{
-		if (!b)
-			return(mp(1,0));
-	PR p=gcd(b,a%b);
-	return(mp(p.y,p.x-p.y*(a/b)));
-}
-
-int Ch(int n,int k)
-{
-	return(fct[n]*ifct[k]%MOD*ifct[n-k]%MOD);
-}
-
-int main()
-{
-	// vars
-	int T,t;
-	int N,Ka,Kb;
-	int i,j,a,b;
-	int ans;
-	// precompute factorials and their modular inverses
-		Fox(i,LIM)
-		{
-			fct[i]=i ? fct[i-1]*i%MOD : 1;
-			ifct[i]=gcd(fct[i],MOD).x;
-				if (ifct[i]<0)
-					ifct[i]+=MOD;
-		}
-	// testcase loop
-	Read(T);
-		Fox1(t,T)
-		{
-			// input
-			Read(N);
-			Read(A[0]),Read(Ka),N=1;
-				while (Ka--)
-				{
-					Read(a),Read(b);
-						Fox(i,b)
-							Read(S[i]);
-						while (a--)
-							Fox(i,b) {
-								A[N]=A[N-1]+S[i];
-								N++;
-							}
-				}
-			Read(B[0]),Read(Kb),N=1;
-				while (Kb--)
-				{
-					Read(a),Read(b);
-						Fox(i,b)
-							Read(S[i]);
-						while (a--)
-							Fox(i,b) {
-								B[N]=B[N-1]+S[i];
-								N++;
-							}
-				}
-			// pad arrays
-			a=-1;
-				Fox(j,4)
-				{
-					A[N+j]=a--;
-					B[N+j]=a--;
-				}
-			// iterate through
-			ans=1;
-			i=j=a=b=0;
-				while ((i<N) || (j<N))
-					if (
-						(A[i]==B[j]) ||
-						(A[i]==B[j+1]) && (A[i+1]==B[j]) ||
-						(A[i]==A[i+2]) && (A[i+1]==B[j]) ||
-						(B[j]==B[j+2]) && (B[j+1]==A[i])
-					) {
-						// cross pattern, must end current streaks
-						ans=(LL)ans*Ch(a+b,a)%MOD;
-						a=b=0;
-							if (A[i]==B[j])
-								i++,j++;
-							else
-							{
-								ans=(ans<<1)%MOD;
-									if (A[i]==A[i+2])
-										i+=3,j++;
-									else
-									if (B[j]==B[j+2])
-										j+=3,i++;
-									else
-										i+=2,j+=2;
-							}
-					}
-					else
-					// continue one of the current streaks?
-					if (A[i]==A[i+1])
-						i+=2,a++;
-					else
-					if (B[j]==B[j+1])
-						j+=2,b++;
-					else
-					if ((A[i]==A[i+2]) && (A[i+1]==A[i+3]))
-						i+=4,a++;
-					else
-					if ((B[j]==B[j+2]) && (B[j+1]==B[j+3]))
-						j+=4,b++;
-					else
-					{
-						// impossible pattern
-						ans=0;
-						break;
-					}
-			ans=(LL)ans*Ch(a+b,a)%MOD;
-			// output
-			printf("Case #%d: %d\n",t,ans);
-		}
-	return(0);
-}
\ No newline at end of file
-- 
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