problem link-click here
it's an easy problem. read the statement carefully then i think you must get a way to solve it.
accepted code is given below:
#include<bits/stdc++.h>
using namespace std;
long long int power(int n,int l)
{
long long int sum=1;
for(int i=1;i<=l;i++)
{
sum=sum*n;
}
return sum;
}
int main()
{
int t;
cin>>t;
while(t--)
{
int a;
vector<int> v;
while(cin>>a && a!=0)
{
v.push_back(a);
}
sort(v.begin(),v.end());
int l=1;
long long int sum=0;
for(int i=v.size()-1;i>=0;i--)
{
sum=sum+(2*power(v[i],l));
l++;
}
if(sum>=5000000)
{
cout<<"Too expensive"<<endl;
}
else
{
cout<<sum<<endl;
}
}
return 0;
}
it's an easy problem. read the statement carefully then i think you must get a way to solve it.
accepted code is given below:
#include<bits/stdc++.h>
using namespace std;
long long int power(int n,int l)
{
long long int sum=1;
for(int i=1;i<=l;i++)
{
sum=sum*n;
}
return sum;
}
int main()
{
int t;
cin>>t;
while(t--)
{
int a;
vector<int> v;
while(cin>>a && a!=0)
{
v.push_back(a);
}
sort(v.begin(),v.end());
int l=1;
long long int sum=0;
for(int i=v.size()-1;i>=0;i--)
{
sum=sum+(2*power(v[i],l));
l++;
}
if(sum>=5000000)
{
cout<<"Too expensive"<<endl;
}
else
{
cout<<sum<<endl;
}
}
return 0;
}
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