Description
Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications:
x2 = x × x, x3 = x2 × x, x4 = x3 × x, …, x31 = x30 × x.
The operation of squaring can be appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications:
x2 = x × x, x3 = x2 × x, x6 = x3 × x3, x7 = x6 × x, x14 = x7 × x7, x15 = x14 × x, x30 = x15 × x15, x31 = x30 × 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, x4 = x2 × x2, x8 = x4 × x4, x8 = x4 × x4, x10 = x8 × x2, x20 = x10 × x10, x30 = x20 × x10, x31 = x30 × x.
If division is also available, we can find a even shorter sequence of operations. It is possible to compute x31 with six operations (five multiplications and one division):
x2 = x × x, x4 = x2 × x2, x8 = x4 × x4, x16 = x8 × x8, x32 = x16 × 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 x for the given positive integer n. Products and quotients appearing in the sequence should be x to a positive integer’s power. In others 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.
题意
将(x)通过乘或除的操作最少多少次能得到(x^n)
题解
设(A[i])为深度i时得到的(x)的次方，最多有(13)次可以到达最大值，从(1)到(13)一次枚举，顺便剪枝：若每次最大次方乘最大次方都无法到达答案，一定不满足，直接返回。
代码
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int n;
int A[20];
int dfs(int d,int mxdep){
if(A[d]==n)return 1;
if(d==mxdep)return 0;
int sz=A[0];
for(int i=1;i<=d;i++)sz=max(sz,A[i]);
if((sz<<(mxdep-d))<n)return 0;
for(int i=d;i>=0;i--){
A[d+1]=A[d]+A[i];
if(dfs(d+1,mxdep))return 1;
A[d+1]=A[d]-A[i];
if(dfs(d+1,mxdep))return 1;
}
return 0;
}
int work(int x){
if(x==1)return 0;
A[0]=1;
for(int i=1;i<13;i++)if(dfs(0,i))return i;
return 13;
}
int main(){
ios::sync_with_stdio(false);
while(1){
cin>>n;
if(n==0)break;
cout<<work(n)<<endl;
}
}