if target > sum(nums[i]) , return -1. Otherwise, an answer exists
提示 2
Solve the problem for each set bit of target, independently, from least significant to most significant bit.
提示 3
For each set bit of target from least to most significant, let X = sum(nums[i]) for nums[i] <= 2^bit.
提示 4
if X >= 2^bit, repeatedly select the maximum nums[i] such that nums[i]<=2^bit that has not been selected yet, until the sum of selected elements equals 2^bit. The selected nums[i] will be part of the subsequence whose elements sum to target, so those elements can not be selected again.
提示 5
Otherwise, select the smallest nums[i] such that nums[i] > 2^bit, delete nums[i] and add two occurences of nums[i]/2. Without moving to the next bit, go back to the step in hint 3.
