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// Hacker Cup 2017
// Round 3
// Broken Bits
// 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);
}
int main()
{
// vars
int T,t;
int N,K;
int i,a,b,ones,origOnes,forceZero;
bool allOnes;
LL v1,v2,ans;
int origL[60],L[60];
// testcase loop
Read(T);
Fox1(t,T)
{
// input
Read(N),Read(K);
ones=0;
Fox(i,N)
Read(L[i]),ones+=L[i];
origOnes=ones=K-ones;
memcpy(origL,L,sizeof(L));
// all unknowns must be 0 or 1?
if ((!ones) || (N==K))
{
ans=0;
goto Done;
}
// 2 leading 0s (can never differentiate them)?
if (L[0]+L[1]==0)
{
ans=-1;
goto Done;
}
// try all possible key 0 sequences
ans=0;
Fox1(a,N-1)
if ((!origL[a]) && (origL[a-1]))
Fox(forceZero,2)
{
// find end of key 0 sequence
memcpy(L,origL,sizeof(L));
ones=origOnes;
FoxI(i,a,N-1)
if (L[i])
break;
b=i-1;
// allocate as many 1s on the left of the key sequence as possible
// (but leaving at least 1, and optionally forcing one of them to be 0)
allOnes=1;
Fox(i,a)
if (!L[i])
if ((ones>1) && ((!allOnes) || (!forceZero)))
L[i]=1,ones--;
else
allOnes=0;
// allocate as many 1s as close to the end as possible (but leaving at least 1)
FoxRI(i,b+1,N-1)
if ((!L[i]) && (ones>1))
L[i]=1,ones--;
// in key sequence, use smallest possible allocation
if (!ones)
continue;
FoxRI(i,a,b)
if (ones)
L[i]=1,ones--;
if (ones)
continue;
// if left of key sequence is all 1s, use 2nd-smallest allocation instead
if (allOnes)
if (!next_permutation(L+a,L+b+1))
continue;
v1=0;
Fox(i,N)
v1=v1*2+L[i];
// state will be known once key sequence wraps around
FoxI(i,a,N-1)
L[i]=1;
v2=0;
Fox(i,N)
v2=v2*2+L[i];
Max(ans,v2-v1+1);
}
// output
Done:;
printf("Case #%d: %lld\n",t,ans);
}
return(0);
}
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