Power Calculus Time Limit: 5000MS Memory Limit: 65536K Total Submissions: 1615 Accepted: 856
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
#include#include using namespace std; int a[2014]; int flag,deep,n; void dfs(int pos) { int t; if(flag||pos>deep) return; if(a[pos]<<(deep-pos) =1 && t<2000) { a[pos+1]=t; dfs(pos+1); } } } int main() { while(cin>>n&&n) { deep=1,flag=0; a[1]=1; while(1) { dfs(1); if(flag) break; deep++; } cout<