Addition Chains

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

 5
 7
 12
 15
 77
 0 

Sample Output

 1 2 4 5
 1 2 4 6 7
 1 2 4 8 12
 1 2 4 5 10 15
 1 2 4 8 9 17 34 68 77

题意：
给你一个数，问你从1开始，最少几个数能加到这个数，其中的一个数（除了1）必须能被其中的两个数相加得到。
思路：既然要最少的，那么这些数越大越好，所以咱们可以倒着搜。
代码：

#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("1n");
            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;
}

最后

以上就是聪慧糖豆最近收集整理的关于Addition Chains的全部内容，更多相关Addition内容请搜索靠谱客的其他文章。