Introduction to competitive programming

Problem :

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
#define rl(x) scanf("%lld",&x)
ll  a1[105];
ll  arr1[2000000];
int main()
{
 ll n,temp1=0,v,ans=0;
 rl(n);
 for(int i=0;i<n;i++)
 rl(a1[i]);
 for(int a=0;a<n;a++)
     if(a1[a])
            for(int b=0;b<n;b++)
        for(int c=0;c<n;c++)
           arr1[temp1++]=(a1[a]*(a1[b]+a1[c]));
        sort(arr1,arr1+temp1);
 for(int d=0;d<n;d++)
           for(int e=0;e<n;e++)
       for(int f=0;f<n;f++)
              {
    v=a1[d]*a1[e]+a1[f];
    ans+=(upper_bound(arr1,arr1+temp1,v))-(lower_bound(arr1,arr1+temp1,v));
              }
 cout<<ans<<endl;
 
} 
 

Problem :

#include<iostream>
using namespace std;
long long phi(long long n) 
{ 
 long  long result = n; 
        for(long  long i=2;i*i <= n;i++) 
        { 
  if (n%i==0) 
  result-=result/i; 
         while (n%i==0) 
  n/=i; 
        } 
        if (n>1) 
 result-=result/n; 
        return result; 
} 
int main()
{
 long long t,n;
 cin>>t; 
 while(t--)
 {
  cin>>n;
  cout<<phi(n)<<endl;
 }
} 

Problem:

The problem is a simple application of BFS,even though the problem can be solved in different methods BFS is easy to implement>
The algorithm is to keep track of all the neighbouring vertices in a queue by checking the boundary cases and incrementing the corresponding distace.Whenever we find the destination point we return the distace at that point

Below code is the c++ implementation for the above problem

 
    #include<set>
    #include<map>
    #include<list>
    #include<stack>
    #include<queue>
    #include<cmath>
    #include<vector>
    #include<cstdio>
    #include<cctype>
    #include<deque>
    #include<bitset>
    #include<utility>
    #include<climits>
    #include<cstdlib>
    #include<cstring>
    #include<iomanip>
    #include<iostream>
    #include<algorithm>
    using namespace std;
    //#include<sai> tHe lAzY PrOgRmMeR :)
    #define rl(T) scanf("%lld",&T)
    #define ri(T) scanf("%d",&T)
    #define rs(T) scanf("%s",T)
    #define rc(T) scanf("%c",&T)
    #define loop(i,a,n) for(long long i=a;i<n;i++)
    #define loopn(i,a,n) for(long long i=1;i<=n;i++)
    #define ll long long
    int x[10]={0,-1,-1,-2,-2,2,2,1,1};
    int y[10]={0,-2,2,-1,1,-1,1,-2,2};
    using namespace std;
    int bfs(int a1,int b1,int a2,int b2)
    {
         int moves[9][9],vis[9][9],m,n;
         pair<int,int>p1;
         queue< pair<int,int> >q;
         p1.first=a1;
         p1.second=b1;
         q.push(p1);
         memset(moves,0,sizeof moves);
         memset(vis,0,sizeof vis);
         vis[a1][b1]=0;
         while(!q.empty())
         {
             p1=q.front();
             q.pop();
             if(p1.first==a2 && p1.second==b2)
             return moves[a2][b2];
             for(int i=1;i<=8;i++)
             {
                 m=p1.first+x[i],n=p1.second+y[i];
                 if(m>8 || m<1 || n>8 || n<1)
                 continue;
                 else
                 {
                     vis[m][n]=1;
                     moves[m][n]=moves[p1.first][p1.second]+1;
                     q.push(make_pair(m,n));
                 }
             }
         }
    }
    int main()
    {
       char a1,a2;
       ll b1,b2,t;
       rl(t);
       while(t--)
       {
          cin>>a1>>b1>>a2>>b2;
          cout<<bfs(a1-'a'+1,b1,a2-'a'+1,b2)<<endl;
       } 
    } 

Problem:

Just find the power of Last disit using Logarithmic exponentiation

Here is the C++ implementation

#include<stdio.h>
long long modexpo(long long a,long long b,long long n)
{
    long long d=1;
    while(b)
    {
        if(b%2)
            d=(d*a)%n;
        b>>=1;
        a=(a*a)%n;
    }
    return d;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long a,b;
        scanf("%lld%lld",&a,&b);
        printf("%lld\n",modexpo(a,b,10));
    }
    return 0;
}

Problem:

The problem is about re-ordering the given trucks in an increasing order by using a temporary street.We can solve it by using stack

Here is the C++ implementation

#include<cstdio>
#include<vector>
#include<stack>
#include<iostream>
using namespace std;
int main()
{
    long long n;
     cin>>n;
     while(n)
    {
        long long ans=1;
 
        long long a[n+10];
        for(long long i=1;i<=n;i++)
            cin>>a[i];
        stack<long long>st;
        long long strt=1;
        for(long long i=1;i<=n;i++)
        {
            if(a[i]==strt)
                strt++;
            else
            {
                if(st.empty())
                {
                      st.push(a[i]);
 
                }
 
                else if(!st.empty() && st.top()<a[i])
                {
 
                    ans=0;
                    break;
                }
                else
                    st.push(a[i]);
            }
 
            while(!st.empty() && st.top()==strt)
            {
                st.pop();
                strt++;
            }
        }
        if(!ans)
            printf("no\n");
        else if(strt-1==n && st.empty())
            printf("yes\n");
            cin>>n;
    }
}

Problem:

This problem can be solved using kmp prefix function

Below is the working code for above problem

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
void prefix(char *s,int n,int *pref)
{
 int x,y;
 int j=0,i;
 pref[0]=0;
 for(int i=1;i<n;i++)
 {
  while(j>0 && s[j]!=s[i])
  j=pref[j-1];
  if(s[i]==s[j]) 
  j++;
  pref[i]=j;
 }
 int ans=1;
 x=n-pref[n-1];
 if(n%x==0)
 ans=n/x;
 printf("%d\n",ans); 
}
int main()
{
 
 char pat[100005];
 scanf("%s",pat);
 while(strcmp(pat,"*")!=0)
 {
 int n=strlen(pat);
 int pref[n+1];
 prefix(pat,n,pref);
 scanf("%s",pat);
 }
} 

Problem:

Here is given a simple problem related to string matching often called as a needle in a haystack

This problem is easily solved with Kmp algorithm

Below is the working code for above problem

#include<cstdio>
#include<cstring>
using namespace std;
char pat[1000010];
char text[1000010];
int pref[1000010];
void prefix(char *s)
{
 int n=strlen(s),flag=1;
 int j=0,i;
 pref[0]=0;
 for(int i=1;i<n;i++)
 {
  while(j>0 && s[j]!=s[i])
  j=pref[j-1];
  if(s[i]==s[j]) 
  j++;
  pref[i]=j;
 }
 
 
}
void search(char *text,char *pat)
{
 int len=strlen(text),flag=1;
 int m=strlen(pat);
 int j=0;
 for(int i=0;i<len;i++)
 {
  while(j>0 && text[i]!=pat[j])
  j=pref[j-1];
  if(text[i]==pat[j])
  j++;
  if(j==m)
  {
  printf("%d\n",i-m+1);
  j=pref[j-1];
  }
 }
 if(flag==1)
        printf("\n");
}
int main()
{
 int n;
 while(scanf("%d\n",&n)!=EOF)
        {
      memset(pref,0,sizeof(pref));
      scanf("%s",pat);
  prefix(pat);
  scanf("%s",text);
  search(text,pat);
 }
 
}