Rank of a Five-Digit Number
National Olympiad — Municipal round
Submission name: number.cpp
Problem summary
A five-digit integer abcde qualifies when b=a+c and d=c+e. Qualifying integers are ordered increasingly and numbered from one. Find the given integer’s position, or zero if it does not qualify.
Input
One five-digit integer.
Output
Its one-based rank, or 0.
Constraints
10000 ≤ input ≤ 99999.
Worked example
An independently prepared example for this explanation.
Input
11000
Output
1
Solution analysis
- Enumerate all five-digit integers up to the input.
- Extract their digits and count those satisfying both equations.
- Return zero unless the input itself satisfies them.
Why it works
Enumeration in numeric order counts precisely all qualifying integers preceding or equal to the input.
Complexity: O(90000) time; O(1) space.
Watch out. Leading zero is forbidden; digit sums are equalities, not arithmetic modulo ten.
Commented C++17 solution
An independently written complete program with standard headers.
// Key idea: Enumeration in numeric order counts precisely all qualifying integers preceding or equal
// to the input.
#include<iostream>
using namespace std;
bool valid(int n){
int e=n%10;
n/=10;
int d=n%10;
n/=10;
int c=n%10;
n/=10;
int b=n%10;
int a=n/10;
return b==a+c&&d==c+e;
}
int main(){
int n;
cin>>n;
if(!valid(n)){
cout<<0<<'\n';
return 0;
}
int rank=0;
for(int i=10000;i<=n;i++){
rank+=valid(i);
}
cout<<rank<<'\n';
}