题目如下：
On a broken calculator that has a number showing on its display, we can perform two operations:
- Double: Multiply the number on the display by 2, or;
- Decrement: Subtract 1 from the number on the display.
Initially, the calculator is displaying the number
X.Return the minimum number of operations needed to display the number
Y.
Example 1:
Input: X = 2, Y = 3Output: 2Explanation: Use double operation and then decrement operation {2 -> 4 -> 3}.Example 2:
Input: X = 5, Y = 8Output: 2Explanation: Use decrement and then double {5 -> 4 -> 8}.Example 3:
Input: X = 3, Y = 10Output: 3Explanation: Use double, decrement and double {3 -> 6 -> 5 -> 10}.Example 4:
Input: X = 1024, Y = 1Output: 1023Explanation: Use decrement operations 1023 times.
Note:
1 <= X <= 10^91 <= Y <= 10^9
解题思路：要增加X只能做乘法操作，要减小X只能做减法。如果X>Y的话，那么只需要一直对X做减法操作直到X=Y为止，operation的次数是X-Y；对于X<Y的情况，我们可以由Y的值反推X，即如果Y是偶数，那么令Y=Y/2，如果Y是奇数，令Y=Y-1，直至Y=X为止。
代码如下：
class Solution(object): def brokenCalc(self, X, Y): """ :type X: int :type Y: int :rtype: int """ if X >= Y: return X - Y res = 0 while Y != X: res += 1 if Y > X and Y % 2 == 0: Y = Y / 2 else: Y = Y + 1 return res