问题 AT: Power Hungry Cow

内存限制:128 MB 时间限制:1.000 S
评测方式:文本比较 命题人:
提交:3 解决:2

题目描述

FJ's cows would like to be able to compute integer powers P(1P20,000) of numbers very quickly, but need your help. Because they're going to be computing powers of very large numbers, they can only keep around two work variables for intermediate results.

The first of those work variables is initialized to the number (denoted x) for which they are calculating the power; the other is initialized to 1. The cows can both multiply and divide any pair of the work variables and store the result in any work variable, but all results are stored as integers.

For example, if they want to compute x31, one way to perform the calculation is:
WV1 WV2

Start: x 1
Multiply first by first, store in second: x x2
Multiply second by second: x x4
Multiply second by second: x x8
Multiply second by second: x x16
Multiply second by second: x x32
Divide second by first: x x31

Thus, x31 can computed in six operations. Given the power to be computed and the the number of work variables, find the minimum number of operations to calculate the power.

输入

A single line with one integer: P.

输出

A single line with a single integer that is the minimum number of operations it requires to compute the power.

样例输入 复制

31

样例输出 复制

6

来源/分类