Power Calculus
Description Starting with x and repeatedly multiplying by x, we can compute x31 with thirty multiplications:
The operation of squaring can be appreciably shorten the sequence of multiplications. The following is a way to compute x31 with eight multiplications:
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:
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):
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 integern. Products and quotients appearing in the sequence 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 Sample Input 1 31 70 91 473 512 811 953 0 Sample Output 0 6 8 9 11 9 13 12 Source |
题意:
算x^n的快速乘法。算过的结果可以利用。乘和除都可以问最少的运算次数。
思路:
由于涉及到最少次数。想用bfs但是。bfs会出问题。因为求解x^n的过程中只能使用已算出的结果。但是bfs不能确定某个状态时那些结果已经算出。所以只能用dfs来求解。但是盲目用dfs每次搜到底的话。时间开销太大。但是我们大概知道需要的运算次数。所以我们可以逐渐加深的来搜这样就可以最快的找到解了。
详细见代码:
#include<algorithm> #include<iostream> #include<string.h> #include<sstream> #include<stdio.h> #include<math.h> #include<vector> #include<string> #include<queue> #include<set> #include<map> //#pragma comment(linker,"/STACK:1024000000,1024000000") using namespace std; const int INF=0x3f3f3f3f; const double eps=1e-8; const double PI=acos(-1.0); const int maxn=100010; typedef __int64 ll; int deep,n,get; int have[1010];//have[i]第i层得到的数 void iddfs(int dep) { int i,tp; if(get||dep>deep||have[dep]<<(deep-dep)<n)//若i次已得到x^m。那么n此处最大能得到x^(m*2^(n-i))。若这个值还小于目标值肯定就不可能了。 return ; if(have[dep]==n) { get=1; return ; } for(i=0;i<=dep;i++) { tp=have[i]+have[dep];//乘 if(tp<2000) { have[dep+1]=tp; iddfs(dep+1); } tp=fabs(have[dep]*1.0-have[i]);//除 if(tp<2000) { have[dep+1]=tp; iddfs(dep+1); } } } int main() { have[0]=1; while(scanf("%d",&n),n) { get=0; deep=-1; while(!get) { deep++;//加深 iddfs(0); } printf("%d\n",deep); } return 0; }