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 | 3985 |
2 | jiangly | 3814 |
3 | jqdai0815 | 3682 |
4 | Benq | 3529 |
5 | orzdevinwang | 3526 |
6 | ksun48 | 3517 |
7 | Radewoosh | 3410 |
8 | hos.lyric | 3399 |
9 | ecnerwala | 3392 |
9 | Um_nik | 3392 |
# | User | Contrib. |
---|---|---|
1 | cry | 169 |
2 | maomao90 | 162 |
2 | Um_nik | 162 |
4 | atcoder_official | 161 |
5 | djm03178 | 158 |
6 | -is-this-fft- | 157 |
7 | adamant | 155 |
8 | awoo | 154 |
8 | Dominater069 | 154 |
10 | luogu_official | 150 |
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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