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hackercup / 2017 /round3 /sluggishsecurity.cpp
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2017 Problems
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// 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);
}