The problem is from SPOJ Online Judge [description here]
Since the input of p and q is restricted to 0 and 1, so we can just check its string value. Bit XOR equals to one if and only if both operands are different.
1 2 3 4 5 6 | s = io.read() if string.sub(s, 1, 1) ~= string.sub(s, 3, 3) then print(1) else print(0) end |
s = io.read() if string.sub(s, 1, 1) ~= string.sub(s, 3, 3) then print(1) else print(0) end
Here, we use string.sub(s, i, j) to get a substring of s starting index i and ending index j. Noted that the index starts at 1 instead of 0.
We can also use bit32.bxor(a, b) to obtain the bit xor for both integers, which you will need to split the input string and convert them into number using tonumber() function.
1 2 3 4 5 6 7 8 9 10 | function split(s, delimiter) result = {}; for match in (s..delimiter):gmatch("(.-)"..delimiter) do table.insert(result, match); end return result; end ss = split(io.read(), " ") io.write(bit32.bxor(tonumber(ss[1]), tonumber(ss[2]))) |
function split(s, delimiter)
result = {};
for match in (s..delimiter):gmatch("(.-)"..delimiter) do
table.insert(result, match);
end
return result;
end
ss = split(io.read(), " ")
io.write(bit32.bxor(tonumber(ss[1]), tonumber(ss[2])))–EOF (The Ultimate Computing & Technology Blog) —
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