algorithm:
if any number(a) is divisible by 5 it can be wriiten as 4*b + b -->
this cleary shows the last two bit of a & b will be same.
lets understand by an example (35)10 = (100011)2
xx1100
+ xx11
---------
100011
now this clearly shows we can calculate the unknowns(x) by traversing
right to left
code:
int main()
{
int n, m;
cin >> n;
m = n;
int a, b;
int i=2;
a = (m&3)<<2;
b = (m&3);
m >>= 2;
bool rem = 0,s,r;
while (m>3)
{
r = a&(1<<i);
s = r^(m&1)^rem;
b = b|(s<<i);
a = a|(s<<(i+2));
rem = (r&s)|(s&rem)|(r&rem) ;
i++;
m >>= 1;
}
if (a+b == n) cout << "yes\n";
else cout << "no\n";
return 0;
}
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