# Daily Coding Challenge #78

Daily Coding Challenge (92 Part Series)

This is a series of Daily Coding Challenge. Each day I show a few solutions written in C++. The questions are from coding practice/contest sites such as HackerRank, LeetCode, Codeforces, Atcoder and etc.

/*
M-SOLUTIONS Programming Contest 2020 - A - Kyu in AtCoder
*/

#include <bits/stdc++.h>
using namespace std;
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

int main()
{
FAST_INP;
int t;
cin >> t;
// straightforward .. follow the Problem Statement
if(t>=400&&t<=599) cout << 8;
else if(t>=600&&t<=799) cout << 7;
else if(t>=800&&t<=999) cout << 6;
else if(t>=1000&&t<=1199) cout << 5;
else if(t>=1200&&t<=1399) cout << 4;
else if(t>=1400&&t<=1599) cout << 3;
else if(t>=1600&&t<=1799) cout << 2;
else if(t>=1800&&t<=1999) cout << 1;
return 0;
}


/*
M-SOLUTIONS Programming Contest 2020 - B - Magic 2
*/

#include <bits/stdc++.h>
using namespace std;
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

int main()
{
FAST_INP;
int a,b,c,k;
cin >> a >> b >> c >> k;
int ok=0;
while(k--){
// The integer on the green card is strictly greater than the integer on the red card.
// The integer on the blue card is strictly greater than the integer on the green card.
// i.e. c > b > a
// if b is less than or equal to a, multiply b by 2
if(b<=a) b*=2;
// if c<=a, then multiply c by 2
// we don't need to multiply a by 2 since it should be the smallest one
else c*=2;
}
// check if it is successful
if(b>a&&c>b) cout << "Yes\n";
else cout <<"No\n";
return 0;
}


/*
M-SOLUTIONS Programming Contest 2020 - C - Marks
*/

#include <bits/stdc++.h>
using namespace std;
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

int main()
{
FAST_INP;
int n,k;
cin >> n >> k;
vector<long long> a(n);
for(int i=0;i<n;i++) cin >> a[i];
for(int i=0;i<n-k;i++){
// we don't need to calculate the grade
// grade for the (K+i)-th term and that for the (K+i-1)-th term have (k-1) common elements
// therefore, just compare a[i] and a[i+k]
if(a[i]<a[i+k]) cout <<"Yes\n";
else cout <<"No\n";
}

return 0;
}

// WA
//int main()
//{
//    FAST_INP;
//    int n,k;
//    cin >> n >> k;
//    vector<long long> a(n), g(n-k+1);
//    for(int i=0;i<n;i++) cin >> a[i];
//    for(int i=0;i<n-k+1;i++){
//      int j=k+i-1;
//      long long sum=1;
//      for(int x=0;x<k;x++){
////            printf("%d * %d \n", sum , a[j-x]);
//          sum*=a[j-x];
//      }
////        printf("%d = %d\n", g[i],sum);
//      g[i] = sum;
//  }
//  for(int i=1;i<n-k+1;i++){
////        printf("%d %d \n", g[i],g[i-1]);
//      if(g[i]>g[i-1]) cout << "Yes\n";
//      else cout << "No\n";
//  }
//    return 0;
//}


/*
M-SOLUTIONS Programming Contest 2020 - D - Road to Millionaire
*/

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
#define FAST_INP  ios_base::sync_with_stdio(false);cin.tie(NULL)

int main()
{
FAST_INP;
int n;
cin >> n;
vector<ll> a(n);
for(int i=0;i<n;i++) cin >> a[i];
ll ans = 1000;
for(int i=0;i<n-1;i++){
// greedy - buy low sell high
ll stock=0;
if(a[i]<a[i+1]) stock = ans/a[i];
ans+=(a[i+1]-a[i])*stock;
}
cout << ans << endl;
return 0;
}


The source code is available in corresponding repo below. Star and watch for timely updates!

## wingkwong / atcoder

### 🏆 A Collection of my AtCoder Solutions with Explanations 🏆

Daily Coding Challenge (92 Part Series)

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### Wing-Kam

Consultant by day. Developer by night. AWS certified. Exploring #CloudNative currently.