Solve a Two-Term Addition Equation
National Olympiad — Municipal round (2016 season)
Submission name: abx.cpp
Problem summary
An equation is either A+X=B or X+A=B, with nonnegative known integers. Find the unknown X, which may be negative.
Input
One equation containing only digits, X, + and =.
Output
The integer value of X.
Constraints
0 ≤ A,B ≤ 1000; no spaces or extra symbols occur.
Worked example
Input and output for the worked example discussed below.
Input
X+12=3
Output
-9
Solution analysis
- Find the + and = positions, then split the equation into the two left-hand terms and the right-hand value B. One of the two terms is X; the other is A.
- Addition is commutative, so both forms give the same equation X=B−A. Parse only the numeric term rather than trying to convert X to an integer.
- In X+12=3, A=12 and B=3, so X=−9. In 12+X=30 the same procedure returns 18. A signed integer is necessary even though both given numbers are nonnegative.
Why it works
Both permitted input forms express A+X=B. Subtracting the known A from both sides gives the unique solution B−A. The delimiter-based parsing identifies exactly those two known values.
Complexity: O(L) time and O(L) temporary string space, where L is the equation length.
Watch out. Do not assume X is always the first term. Permit a negative result and zero-valued known terms. Never apply stoi to the X token.
Commented C++17 solution
An independently written complete program with standard headers.
#include<algorithm>
#include<array>
#include<cmath>
#include<iostream>
#include<limits>
#include<map>
#include<numeric>
#include<set>
#include<string>
#include<vector>
using namespace std;
int main(){
string equation;
cin>>equation;
size_t plus=equation.find('+');
size_t equals=equation.find('=');
string first=equation.substr(0,plus);
string second=equation.substr(plus+1,equals-plus-1);
// Either position of X leads to the same subtraction.
int known=stoi(first=="X"?second:first);
int result=stoi(equation.substr(equals+1));
cout<<result-known<<'\n';
}