【算法•日更•第三十九期】迭代加深搜索：洛谷SP7579 YOKOF - Power Calculus 题解
废话不多说，直接上题：
SP7579 YOKOF - Power Calculus
题意翻译
（略过没有营养的题干）
题目大意： 给出正整数n,若只能使用乘法或除法，输出使x经过运算（自己乘或除自己，以及乘或除运算过程中产生的中间结果）变成x^n的最少步数
输入格式： 若干行数据，每行一个正整数n，数据以单独成行的0结束
输出格式： 若干行数据，对应每行输入的n所需的步数
题目描述
Starting with x and repeatedly multiplying by x, we can compute x 31 with thirty multiplications:
x 2 = x * x, x 3 = x 2 * x, x 4 = x 3 * x, ... , x 31 = x 30 * x.
The operation of squaring can appreciably shorten the sequence of multiplications. The following is a way to compute x 31 with eight multiplications:
x 2 = x * x, x 3 = x 2 * x, x 6 = x 3 * x 3 , x 7 = x 6 * x, x 14 = x 7 * x 7 ,
x 15 = x 14 * x, x 30 = x 15 * x 15 , x 31 = x 30 * x.
This is not the shortest sequence of multiplications to compute x 31 . There are many ways with only seven multiplications. The following is one of them:
x 2 = x * x, x 4 = x 2 * x 2 , x 8 = x 4 * x 4 , x 10 = x 8 * x 2 ,
x 20 = x 10 * x 10 , x 30 = x 20 * x 10 , x 31 = x 30 * x.
There however is no way to compute x 31 with fewer multiplications. Thus this is one of the most efficient ways to compute x 31 only by multiplications.
If division is also available, we can find a shorter sequence of operations. It is possible to compute x 31 with six operations (five multiplications and one division):
x 2 = x * x, x 4 = x 2 * x 2 , x 8 = x 4 * x 4 , x 16 = x 8 * x 8 , x 32 = x 16 * x 16 , x 31 = x 32÷ x.
This is one of the most efficient ways to compute x 31 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 x n by multiplication and division starting with x for 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.
输入格式
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.
输出格式
Your program should print the least total number of multiplications and divisions required to compute x n 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.
输入输出样例
这道题有点尴尬，大多都是英语，幸好浏览器是可以翻译的。
先来确定算法。
这道题先来思考用什么算法？似乎没什么特殊的算法，那么就只能搜索了。
是深搜呢？还是广搜呢？广搜没前途，状态不好记录，深搜又控制不住，一条路走到黑。
其实这道题直接迭代加深搜索就可以了。
什么是迭代加深搜索？就是深搜设定上一个搜索的边界，逐步加深这个边界，这样每次会限制其搜索的深度，就不会一条路走到黑了。
但是这是依旧相当的暴力啊！！！
小编试了一下，连样例数据都卡到爆了，所以必须进一步优化，这里使用剪枝优化。
如果当前的指数自乘剩下的次数之后仍然比n小，那么我们就一定会果断舍弃，这就是剪枝的内容。
代码如下：
1 #include<iostream> 2 using namespace std; 3 int num[10000],n,idt;//用num来存储已经创造过的可以用于计算的数，idt是限制的深度 4 bool dfs(int step,int now)//step是当前用了多少次运算，now是当前指数 5 { 6 if(now<=0||step>idt||now<<(idt-step)<n) return false;//判断一定不能成功的条件和剪枝 7 if(now<<(idt-step)==n) return true;//剪枝 8 if(now==n) return true;//如果正确，那么就返回 9 num[step]=now;//存储一下这个指数 10 for(int i=0;i<=step;i++) 11 { 12 if(dfs(step+1,now+num[i])) return true;//乘 13 if(dfs(step+1,now-num[i])) return true;//除 14 } 15 return false;//不成功一定要最后返回false 16 } 17 int main() 18 { 19 while(1) 20 { 21 cin>>n; 22 if(n==0) break; 23 for(idt=0;;idt++)//从0开始枚举深度 24 if(dfs(0,1)==true) break;//发现可以就结束循环 25 cout<<idt<<endl; 26 } 27 return 0; 28 }
