Accepted solution

463B - Caisa and Pylons

Open original problem
Source code

#include <iostream>

#include<vector>

#include<string>

#include<string.h>

#include<math.h>

#include<set>

#include<map>

#include<list>

#include<stack>

#include<queue>

#include <fstream>

#include<iterator>

#include<algorithm>

#include <iomanip>      // std::setprecision

#include<unordered_set>

/* ==================================================

  |                                                  |

  |    Author: *** S. M. Sabiul Hajjaj Sabit ***     |

  |       Last Edit:  July,25,2019                    |

   ================================================== */

///***global declaration***///

#define all(arr) arr.begin(),arr.end()

#define pb push_back

#define mp make_pair

#define lb lower_bound

#define up upper_bound

#define sz(x) x.size()

#define min3(a,b,c)  min(min(a,b),c)

#define max3(a,b,c)  max(max(a,b),c)

#define lli long long int

#define prnt(x) cout<<x<<endl

#define in1(x) cin>>x

#define in12(x,y) cin>>x>>y

using namespace std;

template <typename T>long long int power(T a, T b)

{

    long long int pow=1;

    for(int i=0; i<b; i++)

        pow *= a;

    return pow;

}

bool isPrime(int n)

{

    int ans=true;

    for(int i=2; i<=n/2; i++)

    {

        if(n%i == 0)

        {

            ans = false;

            break;

        }

    }

    if (ans) return true;

    else return false;

}

/**** ascending sort vector of pair basis fd 2nd element******/

//sort(arr.begin(), arr.end(),sortbysec);

bool sortbysec(const pair<int,int> &a,const pair<int,int> &b)

{

    //return (a.second < b.second); //ascending

    return (a.second > b.second);  //descending

}

/////*********Write Code From Here*********///

int main()

{

    int n;

    in1(n);

    int arr[99999];

    for(int i=1; i<=n; i++)

    {

        in1(arr[i]);

    }

    int ans =0,t=0;

    for(int i=1; i<=n; i++)

    {

        t += (arr[i-1]-arr[i]);

        if(t<0)

        {

            ans+= -(t);

            t=0;

        }

    }

    prnt(ans);

}
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