[TargetLowering] Add support for more constants to expandDIVREMByConstant. (#189286) If we can find a chunk size k where 2^k mod divisor == -1, we can add the even chunks and subtract the odd chunks. The resulting sum may be negative, so we need to add 2^k + 1 (a multiple of divisor) for each odd chunk to ensure the result is positive. We have enough extra bits between the chunk size and HBitWidth to avoid overflow. The idea here is similar to checking if a decimal number is divisible by 11. You can add the even digits and subtract the odd digits. If the resulting sum is divisible by 11 the original number is divisible by 11.
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