TechoSagar: Numbers containing 0's from 1 to N

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Numbers containing 0's from 1 to N


Efficiently count how many integers from 1 to N contains 0's as a digit. 
Input:
First line of the input contains the number of test cases T. Each line of test case contains the integer N.

Output:
Number of integers that contain 0 as a digit are printed.

Constraints:
1 <=T<= 100
1 <=N<= 10000

Example:
Input:

3
100
987
20
Output:
10
179
2

Your Code
#include<iostream>

using namespace std;

#include<sstream>

int main()
{
int t;
cin>>t;
while(t--)
{ int n;
cin>>n;
int i,sum=0;
for(i=1; i<=n; i++)
{    int temp=0;
stringstream ss;
ss<<i;
string str=ss.str();
int l;
l=str.length();
int j;
for(j=l-1; j>=0; j--)
{
if(str[j]=='0')
{
temp=1;
break;
}
}
if(temp==1)
sum++;
}
cout<<sum<<endl;
}
}

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