## A - Bitwise Exclusive Or Editorial by en_translator

If you are new to learning programming and do not know where to start, please try Problem A “Welcome to AtCoder” from practice contest. There you can find a sample code for each language.

#### Using the property of XOR

If the theme is an addition instead of XOR, that is, if the problem is “find \(C\) such that \(A+C=B\),” then we can subtract \(A\) from both sides to obtain \(C=B-A\). What is the operation in XOR corresponding to “subtracting \(A\) from both sides” in the case of addition?

Actually, for any non-negative integer \(A\), we have \(A \ XOR \ A=0\). Therefore, by XOR-ing \(A\) to both sides, the \(A\) in the left hand side is eliminated, and now we have \(C=B \ XOR \ A\) that enables us to calculate \(C\).

In many languages, the `^`

operator is served for XOR computation.

Sample code (C)

```
#include<stdio.h>
int main(){
int a,b;
scanf("%d%d",&a,&b);
printf("%d\n",a^b);
}
```

Sample code (Python)

```
A,B = map(int, input().split())
print(A^B)
```

#### Brute-forcing

We know that the answer is one of the \(256\) integers in the range of \(0\) and \(255\). We can solve the problem by checking all the \(256\) cases if the equivalence is satisfied.

When implementing, you have to be careful of the precedence of operators.

Sample code (C)

```
#include<stdio.h>
int main(){
int a,b;
scanf("%d%d",&a,&b);
for(int c=0;c<=255;c++){
if((a^c)==b){
printf("%d\n",c);
return 0;
}
}
}
```

Sample code (Python)

```
A,B=map(int,input().split())
for C in range(256):
if (A^C)==B:
print(C)
exit()
```

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