rahulkumarsk2015
BAN USERUsing kmp algorithm we can solve in n^3
- rahulkumarsk2015 May 30, 2018First calculate the root node... Then recursively call for left and right child to make complete bst
- rahulkumarsk2015 May 29, 2018It can be done in m*n*n using kadane algorithm.. Taking each pairs of column and try to apply the kadane in every row
- rahulkumarsk2015 May 28, 2018#include<bits/stdc++.h>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,i,j,cnt=1;
cin>>n;
char p[n];
string str[n];
for(i=0;i<n;i++)
cin>>p[i];
for(i=0;i<n;i++)
cin>>str[i];
for(i=0;i<n-1;i++)
{
for(j=i+1;j<n;j++)
{
if(p[i]==p[j])
{
if(str[i]!=str[j])
{
cnt=0;
}
}
if(p[i]!=p[j])
{
if(str[i]==str[j])
{
cnt=0;
}
}
if(cnt==0)
{
cout<<"n0";
break;
}
}
if(cnt==0)
{
break;
}
}
if(cnt==1)
{
cout<<"yes";
}
}
}
#include<iostream>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,i,j,k,l;
cin>>n>>k>>l;
float ar1[n+1][n+1];
for( i=0;i<=n;i++)
for(j=0;j<=n;j++)
{
ar1[i][j]=0;
}
ar1[1][1]=n;
for(i=2;i<=n;i++)
{
for(j=1;j<=i-1;j++)
{
ar1[i][j]=(ar1[i-1][j]-1)*1.0/2;
ar1[i][j+1]=(ar1[i-1][j]-1)*1.0/2;
}
}
if(ar1[k][l]>0 && ar1[k][l]<=1)
cout<<ar1[k][l];
if(ar1[k][l]>1)
cout<<"1";
else
cout<<"0";
}
}
#include<iostream>
using namespace std;
int main()
{
int t;
cin>>t;
while(t--)
{
int n,i,j,k,l;
cin>>n>>k>>l;
float ar1[n+1][n+1];
for( i=0;i<=n;i++)
for(j=0;j<=n;j++)
{
ar1[i][j]=0;
}
ar1[1][1]=n;
for(i=2;i<=n;i++)
{
for(j=1;j<=i-1;j++)
{
ar1[i][j]=(ar1[i-1][j]-1)*1.0/2;
ar1[i][j+1]=(ar1[i-1][j]-1)*1.0/2;
}
}
if(ar1[k][l]>0 && ar1[k][l]<=1)
cout<<ar1[k][l];
else
cout<<"0";
}
}
We can use dynamic programming... Dp[I] will store the total number from last position.. Dp[i+1]=dp[i]+dp[i-2];
- rahulkumarsk2015 May 30, 2018