Codeforces Problem 102B Solution in (C++)

Welcome:

The question to the problem can be found here:

Explanation:

1.Enter the integer as an character.

2.This will separate the biggest integer into smaller pieces.

3.Convert the pieces into integer and then add using a variable and store in the array s.

4.Convert the piece into character array

5.Continue until only 1 character is present.

6.Happy Coding.

Code:

#include<bits/stdc++.h>

using namespace std;

char s[100000];

int main(){

int cnt=0;

cin>>s;

while(s[1]){

cnt++;

int res =0;

for(int i=0;s[i];i++)

    res+= s[i] - '0';

    sprintf(s,"%d",res);

}

cout<<cnt;

return 0;}

Hackerrank Problem Breadth First Search: Shortest Reach C++

Welcome:

The question to the problem can be found here:

Explanation:
1.Take input as explained in the question.
2.Conduct a breadth first search taking s(input) as the starting index and n(last node).
3.Print all the distances available from s. If level is  zero print -1 that means there is no path from the start symbol to this node.

Happy Coding :)

Code:

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

vector<int> adj[10000]; //adj[a].push_back(b); for add an edge from a to b
int visited[10000]={0}; //O if not visited, 1 if visited
int level[10000];

void addEdge(int v, int w)
{
    adj[v].push_back(w); // Add w to v’s list.
 adj[w].push_back(v);
}

void bfs(int s, int n)
{
    for(int i=0; i<10000;i++){
        visited[i] = 0;
    level[i]=0;
    }
    queue<int>Q;
    Q.push(s);
    visited[s] = 1;
    level[s] = 0;

    while(!Q.empty())
    {
        int u = Q.front();

        for(int i=0; i<adj[u].size(); i++){
            if(visited[adj[u][i]]==0){
                int v = adj[u][i];
                level[v] = level[u]+6;
                visited[v] = 1;
                Q.push(v);
            }
        }
        Q.pop();
    }

  for(int i=1;i<=n;i++)
    {
        if(i!=s)
        {
            if(level[i]==0)
            cout<<"-1 ";
            else
            cout<<level[i]<<" ";
        }
    }
    cout<<endl;
}

int main()
{
    int q;
   cin>>q;
  while(q-->0){

    int n,m;
   cin>>n>>m;
    for(int i=0;i<n+3;i++)
         adj[i].clear();
   for(int i=0;i<m;i++){
        int u,v;
   cin>>u>>v;
   addEdge(u,v);
   }
   int s;
   cin>>s;
bfs(s,n);
}
}

Codeforces Problem 864A Solution in (C++)

Welcome:

The question to the problem can be found here:

Explanation:
1. a[entered input] is the location of the input, say 10 it means the element is in the 10th position in the array.
2. Increase the value in that position by 1 when you get a different match of integers.
3. Store the first identical element in the 1st place of another array com[0] , second element to com[1]
4. Print YES if there are at exactly 2 different number in all cases.Then print those two numbers.
5. If not Print NO
6. Happy Coding!

Code:
#include<bits/stdc++.h>
using namespace std;
int a[100], com[2];
main(){
    int sto=0;
int n;
cin>>n;
int x;
for(int i=0;i<n;i++){
cin>>x;
cin.ignore();
if(++a[x] == 1){ //if more than 1 loop condition becomes false;
    com[sto++] = x;
  
}
}

if(sto == 2 && a[com[0]]== a[com[1]]){
    cout<<"YES"<<endl;
    cout<<com[0]<<" "<<com[1];
}
else{
    cout<<"NO"<<endl;
}

}

Finding Array of Prime and Factors using Sieve of Eratosthenes Algorithm in C++

Theory: Using the sieve of Eratosthenes find an array of prime numbers.

Explanation:

1.Declare an int array sieve[]
2.Using the Sieve of Eratosthenes algorithm an array of prime can be obtained.Again factors of non prime can also be obtained;
3.sieve[i] == 0 : i is a prime number.
4. When not equal to zero the factor is displaced within the array element.

Code
#include <bits/stdc++.h>
using namespace std;
int sieve[100];
int main(){
int n;

cin>>n;
for(int x=2; x<=n; x++){
if(sieve[x]){
continue;
}
for(int i=2*x;i<=n;i+=x){
sieve[i]=x;
}
}

cout<<"Prime factors is those elements with 0 and other elements are factors";
for(int i=2;i<=n;i++){

    cout<<i<<" "<<sieve[i]<<endl;
}
return 0;}

CodeForces Problem 863B - Kayaking Solution (C++)

Problem:

Please find the problem here.

Solution:

This is the easiest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include<bits/stdc++.h>
using namespace std;
int a[100],b[100];
main(){
    int n;
    cin>>n;
    for(int i=0;i<2*n;i++){
        cin>>a[i];
    }
    int maximum = 1000000;
    for(int i=0;i<2*n;i++){
        for(int j=i+1;j<2*n;j++){
            for(int k=0;k<2*n;k++){
             b[k]=a[k];
            }

        b[i]=b[j]= -1;
        sort(b,b+2*n);
             int ans=0;

        for(int k=2;k<2*n-1;k+=2){
           ans+= b[k+1]- b[k];
        }

        maximum=min(maximum,ans);
        }

    }

    cout<<maximum;




}

CodeForces Problem 862A - Mahmoud and Ehab and the MEX Solution (C++)

Problem:

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Explanation:

1.MEX has two condition

2.MEX equal to entered input ... MEX will increase by 1 as element is erased.

3.MEX less than input...MEX will decline as element is inserted before MEX

4.Goodluck!

Code:

#include<bits/stdc++.h>

#define FOR(i,n) for(int i=0;i<n;i++)

using namespace std;

main(){

int n;

int k;

int x;

cin>>n>>k;

int ans=k;

FOR(i,n){

cin>>x;

if(x<k){

ans--;

}

else if(x==k){

ans++;

}

}

cout<<ans;

}

Sum of Maximum SubArray O(n) Time Complexity Short Code (C++)

//Find the maximum sum of subarray
#include <bits/stdc++.h>
using namespace std;
int main(){
int n;
n=8; //size of array
int arr[] ={-2,-3,4,-1,-2,1,5,-3};
int best= 0, sum =0;
for(int i=0;i<n;i++){
sum = max(arr[i],sum+ arr[i]);  // max of array
best= max(best,sum);  // Largest sum of subarray

}
cout<< best<<"\n";


return 0;
}

Codeforce Problem 854A - Fraction Solution (C++)

Problem:

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>

using namespace std;
int main(){
int n,sum;
cin>> n;
for(int i=n/2;i>=1;i--){
    if(__gcd(i,n-i)== 1){

        cout<<i<< " "<<n-i<<endl;
return 0;
    }

}

}

UVa Problem 594 - One Little, Two Little, Three Little Endians Solution (C++)

 Problem:

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Hint: swaping the character pointer for each bits reverses the entire byte. And displaying the value after reversal.

Code:

#include<bits/stdc++.h>

using namespace std;

#define swp(a,b) a=a^b, b=a^b, a=a^b

main(){

int n,reverse;

while(cin>>n){

reverse =n;

char* bits =(char*) &reverse;

swp(bits[0],bits[3]);

swp(bits[1],bits[2]);

cout<<n<<" converts to "<<reverse<<endl;

}

}

UVa Problem 483 - Word Scramble (C++)

Problem:

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <string>

#include <iostream>

using namespace std;

int main(){

    string line;

    while(getline (cin,line)) {                          

                  int tam = line.size();//10 size of first line

                  int cont;

                  string sub;

                  for(int i=0;i<tam;i++){    //I

                       cont=0;

                       if(line[i]!=' '){   // If there is no space

                                while(line[i+cont]!=' ' && i+cont<tam){

                                       cont++; //count becomes 1

                                }

                                for(int j=(i+cont)-1;j>=i && j<tam;j--){ //print last word j then decrease j--

                                          cout<<line[j];

                                }

                                i+=cont-1; //the next word which is a space

                       }else{ //if there is space

                            cout<<line[i];

                       }

                  }

                  cout<<endl;

    }

 return 0;

}

Uva Problem 11936 - The Lazy Lumberjacks Solution (C++)

Problem:

11936 - The Lazy Lumberjacks

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>

using namespace std;

int main(){

int t;

int x,y,z;

cin>>t;

while(t--){

cin>>x>>y>>z;

if(x+y>z && y+z>x && z+x>y){

cout<<"OK"<<endl;

}

else if(x==y && y ==z && z==x)

    cout<<"OK"<<endl;

else cout<<"Wrong!!"<<endl;

}

return 0;}

OjLight Problem 1294 - Positive Negetive Number (C++)

Problem:

Solution:

This is the one of the simplest problem I ever had, just implement using the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>
using namespace std;
int main(){
long long t;
long long n,m,result;
cin>>t;
for(int i=1;i<=t;i++){
    cin>>n>>m;
    result = m*(n/2);
    cout<<"Case "<<i<<": "<<result<<endl;

}
return 0;

}

OjLight Problem 1053 - Higher Math Solution (C++)

Problem: OjLight Problem 1053 - Higher Math

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>

using namespace std;

main(){

long long n,a,b,c,f,g;

cin>>n;

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

        cin>>a>>b>>c;

if(a>b && a>c)

{

    f= a*a;

    g=(b*b)+(c*c);

}

else if(b>a&&b>c){

    f= (b*b);

    g= a*a+c*c;

}

else if(c>a&& c>b){

    f=c*c;

    g=a*a+b*b;

}

(f==g?cout<<"Case "<<i<<":"<<" yes"<<endl:cout<<"Case "<<i<<":"<<" no"<<endl);

}

return 0;

}

CodeForce Problem 588A - Duff and Meat Solution (C++)

Problem:

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

The code is explained below using Comments(//)

//http://codeforces.com/problemset/problem/588/A

#include <bits/stdc++.h>
using namespace std;
#define REP(i,a,b) for(int i=a;i<=b;i++)            // Loop starts from a and ends at b
main(){
int n,a,p;
cin>>n;                                                               //Input taken
int mi =200;
int cost=0;
REP(i,0,n-1){
cin>>a;
cin>>p;
mi=min(mi,p);                                                     //Finding the minimum and replacing the minimum.
cost =cost+(a*mi);                                              // Multiplying minimum with required amount .
}                                                                          // Add to Cost
cout<<cost<<endl;    
}

CodeForce Problem 318A - Even Odds (C++)

Problem:

Please find the problem here.

Solution:

This is one of the easiest problem I ever had, just implement the given formulas! However, it does take me a few minutes to write.

Code:

//Even Odds Solution
#include <bits/stdc++.h>
#define REP(i,a,b) for(int i=a;i<=b;i++)
typedef long long ll;
using namespace std;
int main(){
ll n, k;
cin>>n>>k;
//Suppose n=10.
//Odd series then Even series 1,3,5,7,9,2,4,6,8,10.
n=(n+1)/2; //Finding the mid point.In above case it is 9.
//Condition 1.greater than 9 Even and vice-versa(Odd).
cout<<(k>n? 2*(k-n):2*k-1)<<endl;//if and else condition

return 0;
}

CodeForce Problem 849A - Odds and Ends Solution (C++)

Problem:

You can see the question to the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>

using namespace std;

main(){

int n;

int s[101];

cin>>n;

for(int i=0;i<n;i++){

    cin>>s[i];   //Numbers are stored into array

}

/*Algorithm explanation

//when n is divided by 2 remainder is not 0, result is always odd.

//Plus First element and last element.. both of them must always be odd for this Algorithm to work as mentioned in question. */

(n%2 && s[0]%2 && s[n-1]%2)?cout<<"Yes":cout<<"No";  //Another form of if else.

}

UVa Problem 272 - TEX Quotes Solution (C++)

Problem:

Please find the problem here.

Solution:

This is the simplest problem I ever had, just implement the given formulas! However, it does take me some time to write.

Code:

#include <bits/stdc++.h>
using namespace std;
main(){
    
    int c =0;
string s;
while(getline(cin,s)  ){
    while(s.find('"')+1){
    if(c==0  ){
        s.replace(s.find('"'),1,"``");
        c=1;
    }
    else{
             s.replace(s.find('"'),1,"''");
        c=0;
    }
    }
    cout<<s<<endl;
      
}




}