I'm interested in whether we can Construct a tree if we know the Starting Time and Ending Time of all the Vertices.
For Example:
- Number of Vertices :4
- Starting time ST[i]:2 4 1 3
- Ending time FT[i]:2 4 4 4
The Tree Looks Like:
# | User | Rating |
---|---|---|
1 | tourist | 3843 |
2 | jiangly | 3705 |
3 | Benq | 3628 |
4 | orzdevinwang | 3571 |
5 | Geothermal | 3569 |
5 | cnnfls_csy | 3569 |
7 | jqdai0815 | 3530 |
8 | ecnerwala | 3499 |
9 | gyh20 | 3447 |
10 | Rebelz | 3409 |
# | User | Contrib. |
---|---|---|
1 | maomao90 | 171 |
2 | awoo | 163 |
3 | adamant | 162 |
4 | TheScrasse | 158 |
5 | nor | 153 |
5 | maroonrk | 153 |
7 | -is-this-fft- | 152 |
8 | Petr | 146 |
9 | orz | 145 |
10 | pajenegod | 144 |
I'm interested in whether we can Construct a tree if we know the Starting Time and Ending Time of all the Vertices.
For Example:
The Tree Looks Like:
Hello everyone,
I was working on a problem where i was given an integer p and was asked to calculate the number of digits after the decimal point in 1/p.
Ex:- Input: 100 Output: 2 as (1/100=0.01)
#include<bits/stdc++.h>
using namespace std;
int main()
{
float p;
cin>>p;
float q=1/p;
int cnt=0;
cout<<q<<"\n";
while(fmod(q,10.0)!=0)
{
q=q*10;
cnt++;
//cout<<q<<" "<<cnt<<"\n";
}
cout<<cnt-1<<"\n";
}
What i was trying to do is multiply (1/n) till it is divisible 10 and then simply print count-1 .
But the output this code is giving me is 12 in place of 2 for p=100.
Help would be appreciated :)
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