uva1374-Power Calculus
出處 ： https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4120
Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications:
x2 = x x x, x3 = x2 x x, x4 = x3 x x, ... , x31 = x30 x x.
The operation of squaring can appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications:
x2 = x x x, x3 = x2 x x, x6 = x3 x x3, x7 = x6 x x, x14 = x7 x x7,
x15 = x14 x x, x30 = x15 x x15, x31 = x30 x x.
This is not the shortest sequence of multiplications to compute x31. There are many ways with only seven multiplications. The following is one of them:
x2 = x x x, x4 = x2 x x2, x8 = x4 x x4, x10 = x8 x x2,
x20 = x10 x x10, x30 = x20 x x10, x31 = x30 x x.
There however is no way to compute x31 with fewer multiplications. Thus this is one of the most efficient ways to compute x31 only by multiplications.
If division is also available, we can find a shorter sequence of operations. It is possible to compute x31 with six operations (five multiplications and one division):
x2 = x x x, x4 = x2 x x2, x8 = x4 x x4, x16 = x8 x x8, x32 = x16 x x16, x31 = x32 ÷ x.
This is one of the most efficient ways to compute x31 if a division is as fast as a multiplication.
Your mission is to write a program to find the least number of operations to compute xn by multiplication and division starting with xfor the given positive integer n. Products and quotients appearing in the sequence of operations should be x to a positive integer's power. In other words, x-3, for example, should never appear.
Input
The input is a sequence of one or more lines each containing a single integer n. n is positive and less than or equal to 1000. The end of the input is indicated by a zero.
Output
Your program should print the least total number of multiplications and divisions required to compute xn starting with x for the integer n. The numbers should be written each in a separate line without any superfluous characters such as leading or trailing spaces.
Sample Input
1
31
70
91
473
512
811
953
0
Sample Output
0
6
8
9
11
9
13
12
解題策略
使用iterative deepening，不斷增加深度
每次只能使用一個乘法或除法，一次只能增加一個新的次方
bool dfs(int d,int p){//d為目前深度，p為新產生的次方值
利用v[]紀錄是否已經拜訪過
利用ans[]紀錄每個深度的次方
若最後剩餘的maxd-d，都使用x^p、x^p^2、x^p^4...x^p^2^(maxd-d) 都無法到target就結束dfs，return false