Addition Chains

An addition chain for n is an integer sequence <a0, a1,a2,…,am> with the following four properties:

a0 = 1
am = n
a0 < a1 < a2 < … < am-1 < am
For each k (1 <= k <= m) there exist two (not necessarily different) integers i and j (0 <= i, j <= k-1) with ak = ai + aj

You are given an integer n. Your job is to construct an addition chain for n 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

The input will contain one or more test cases. Each test case consists of one line containing one integer n (1 <= n <= 100). Input is terminated by a value of zero (0) for n.

Output

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

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

題意：
給你一個數(shù)，問你從1開始，最少幾個數(shù)能加到這個數(shù)，其中的一個數(shù)（除了1）必須能被其中的兩個數(shù)相加得到。
思路：既然要最少的，那么這些數(shù)越大越好，所以咱們可以倒著搜。
代碼：

#include<stdio.h> #include<string.h> int a[110]; int s,c[110],n; void dfs(int x,int b) {if(x>n||b>=s)return;if(x==n&&b<s){s=b;for(int i=0; i<s; i++)c[i]=a[i];return;}for(int i=b-1; i>=0; i--){a[b]=x+a[i];dfs(a[b],b+1);a[b]=0;}return; } int main() {while(~scanf("%d",&n),n){memset(a,0,sizeof(a));memset(c,0,sizeof(c));s=110;if(n==1){printf("1\n");continue;}a[0]=1,a[1]=2;dfs(2,2);printf("%d",c[0]);for(int i=1;i<s;i++)printf(" %d",c[i]);printf("\n");}return 0; }

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