uva 529 - Addition Chains

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Addition Chains

An addition chain for n is an integer sequence $<a_0, a_1, a_2, \dots, a_m>$ with the following four properties:

a₀ = 1
a_m = n
a₀<a₁<a₂<...<a_m-1<a_m
For each k ( $1 \le k \le m$ ) there exist two (not neccessarily different) integersi and j ( $0 \le i, j \le k-1$ ) witha_k =a_i +a_j

You are given an integer n. Your job is to construct an addition chain forn with minimal length. If there is more than one such sequence, any one is acceptable.

For example, <1,2,3,5> and <1,2,4,5> are both valid solutions when you are asked for an addition chain for 5.

Input Specification

The input file will contain one or more test cases. Each test case consists of one line containing one integern ( $1 \le n \le 10000$ ). Input is terminated by a value of zero (0) forn.

Output Specification

For each test case, print one line containing the required integer sequence. Separate the numbers by one blank.

Hint: The problem is a little time-critical, so use proper break conditions where necessary to reduce the search space.

Sample Input

571215770

Sample Output

1 2 4 51 2 4 6 71 2 4 8 121 2 4 5 10 151 2 4 8 9 17 34 68 77

迭代加深dfs很适合n=10000，17个数字，虽然感觉上迭代本身存在大量重复计算，但是重复的搜索树的深度小；

迭代加深搜索算法就是仿广度优先搜索的深度优先搜索。既能满足深度优先搜索的线性存储要求，又能保证发现一个最小深度的目标结点。

#include <string.h>#include <stdio.h>int a[100],best[100],n,min,f;int minlen(int num,int len){int x=num,y=len; while (x<n) {x=x<<1;++y;} return y;}void dfs(int len){ int i,j,time,x; if ((a[len]==n)&&(len<min)) {f=0; min=len; for (i=1;i<=len;i++) best[i]=a[i]; } if ((a[len]>=n)||(len>=min)) return ; if (minlen(a[len],len)>=min) return ; for (i=1;i<=len;i++) for (j=i;j<=len;j++) {  a[len+1]=a[i]+a[j];  if (a[len+1]>a[len]) dfs(len+1); }}int main(){ int i; a[1]=1; a[2]=2; best[1]=1; best[2]=2; while (scanf("%d",&n),n) {  f=1; min=minlen(2,2)-1;  if (n<=2) {min=n;f=0;}  while (f)  {++min;   dfs(2);  }  for (i=1;i<min;i++)     printf("%d ",best[i]);  printf("%d\n",n); } return 0;}