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hackercup / 2018 /round3 /ethan_max_subarray.cpp
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2018 Problems
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// Ethan Finds the Maximum Subarray Sum
// Solution by Jacob Plachta
#define DEBUG 0
#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); }
string plural(string s) { return(Sz(s) && s[Sz(s)-1]=='x' ? s+"en" : s+"s"); }
const int INF = (int)1e9;
const LD EPS = 1e-12;
const LD PI = acos(-1.0);
#if DEBUG
#define GETCHAR getchar
#else
#define GETCHAR getchar_unlocked
#endif
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()
{
if (DEBUG)
freopen("in.txt","r",stdin);
// vars
int T,t;
int N,K;
int i,m,w,s,c,ans;
int A[100];
// testcase loop
Read(T);
Fox1(t,T)
{
// input
Read(N),Read(K);
N=N*2-1;
m=0;
Fox(i,N)
if (i%2==0)
Read(A[i]),Max(m,A[i]);
if (A[N-1]<0)
N--;
// consider each possible value w for Ethan's answer
ans=0;
FoxI(w,m,N*K+1)
{
// greedily fill in values
s=c=0;
FoxI(i,(A[0]<0 ? 1 : 0),N-1)
if (i%2==0) // forced value?
{
s+=A[i];
if (A[i]>=0)
c+=A[i];
else
c=0;
}
else
if (i==N-1) // last value?
{
if (c+K<=w) // fill with K if possible
s+=K,c+=K;
}
else
if (((A[i+1]<0) && (c+K<=w)) || ((A[i+1]>=0) && (c+K+A[i+1]<=w))) // can fill with K?
s+=K,c+=K;
else
s--,c=0; // fill with -1 otherwise
Max(ans,s-w);
}
// output
printf("Case #%d: %d\n",t,ans);
}
return(0);
}