For each unsigned 32-bit number, print how many bits are set to 1, and whether the number is an exact power of two.
Use bitwise operators, not division or pow().
Constraints
- 1 ≤ n ≤ 100000
- 0 ≤ each number ≤ 2^32 − 1
Input and output
- Input
- First line: n. Second line: n numbers, each between 0 and 4294967295.
- Output
- n lines. Each line has the set-bit count, a space, then `yes` if the number is a power of two (1, 2, 4, 8, …) or `no` otherwise.
Examples
Input 1
4
7 8 0 4294967295Output
3 no
1 yes
0 no
32 no7 = 111 (3 bits). 8 = 1000 is a power of two. 0 has no set bits. 2^32 − 1 has all 32 bits set.
Input 2
3
1 6 1024Output
1 yes
2 no
1 yes1 = 2^0 and 1024 = 2^10 are powers of two; 6 = 110 is not.
Show hints
x & (x - 1)clears the lowest set bit of x.- A power of two has exactly one set bit. Is 0 a power of two?
Show the explanation
Kernighan's loop. x & (x - 1) turns off the lowest 1-bit: subtracting 1 flips that bit and every 0 below it, and the AND keeps only the higher bits. Repeat until x is 0 and count the iterations. The loop runs once per set bit, not once per bit position.
Power of two. A power of two has exactly one set bit, so it satisfies x != 0 && (x & (x - 1)) == 0. The x != 0 part matters: 0 has no set bits and is not a power of two.
Types. Use unsigned int (or uint32_t) and read with %u. Values up to 4294967295 do not fit in a signed 32-bit int, and bit tricks on negative signed numbers are a classic source of bugs. GCC and Clang also offer __builtin_popcount, which compiles to a single instruction on most CPUs. Mention it, but show the loop first.
Original UniCircle question. Last reviewed 22 Oct 2026.