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.
Sample Input
1 31 70 91 473 512 811 953 0
Sample Output
0 6 8 9 11 9 13 12
Source
求只用乘法和除法最快多少步可以求到x^n
其实答案最大13，但由于树的分支极为庞大在IDDFS的同时，我们还要加2个剪枝
1 如果当前序列最大值m*2^(dep-k)<n则减去这个分支
2 如果出现两个大于n的数则要减去分支。因为里面只有一个有用，我们一定可以通过另外更加短的路径得到答案
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 using namespace std; 5 int num; 6 int way[1006]; 7 bool dfs(int n,int step){ 8 if(num>step) return false; 9 if(way[num]==n) return true; 10 if(way[num]<<(step-num)<n) return false;//强剪枝 11 for(int i=0;i<=num;i++){ 12 num++; 13 way[num]=way[num-1]+way[i]; 14 if(way[num]<=1000 && dfs(n,step)) return true; 15 16 way[num]=way[num-1]-way[i]; 17 if(way[num]>0 && dfs(n,step)) return true; 18 num--; 19 } 20 return false; 21 } 22 int main() 23 { 24 int n; 25 while(scanf("%d",&n)==1){ 26 if(n==0){ 27 break; 28 } 29 30 //迭代加深dfs 31 int i; 32 for(i=0;;i++){ 33 way[num=0]=1; 34 if(dfs(n,i)) 35 break; 36 } 37 printf("%d\n",i); 38 39 } 40 return 0; 41 }
poj 3134 Power Calculus（迭代加深dfs+强剪枝）
原文：http://www.cnblogs.com/UniqueColor/p/4776943.html
