After n jumps, if you jump always to the right, you'll be at point p = 1 + 2 + 3 + 4 + ... + n. If, instead of jumping right, you jumped left in the k_th jump, you would be at point p - 2k. Moreover, by carefully choosing which jumps to go left and which to go right, after n jumps, you can be at any point between n * (n + 1) / 2 and  - (n * (n + 1) / 2) with the same parity as n * (n + 1) / 2. So, what you must do is simulate the jumping process, always jumping to the right, and if at some point, you've reached a point that has the same parity as x and is at or beyond x, you'll have your answer.