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.
Hint: The problem is a little time-critical, so use proper break conditions where necessary to reduce the search space.
Hint: The problem is a little time-critical, so use proper break conditions where necessary to reduce the search space.
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:x【1】 = 1，x【m】 = n，给一个n，求出m最小的序列并输出 思路：这种最短的问题很容易直接想到bfs，但是写法可能不是很经常遇到，网上也有bfs写法的代码，就不给出代码了，主要就是bfs的路径记录，用个vector储存路径，并且每个路径记录前置路径，当达到答案n时，路径肯定最短，然后沿着记录的前置路径递归
其次可以有dfs，dfs的话有两种，一种是迭代加深，就是枚举搜索深度，如果没搜索到答案，就增加深度重新搜索，这种方法主要用在确定答案可以在较小深度的情况，时间复杂度类似bfs，但是空间复杂度小
最后就是dfs的剪枝版本了，剪枝的话其实是一个最优性剪枝（极限思想）和搜索顺序，搜索顺序正常从大到小，我们发现从打一个数x1递增到x2，所需要的最小步数是>=log（x2-x1），这样如果当前步数+log（x2-x1） > 当前记录的答案，这就说明这一定不是最优 迭代加深：
#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std; int ans[];
bool vis[];
int n;
bool dfs(int last,int lim)
{
bool vis[];
memset(vis,,sizeof(vis));
if(last == lim)
{
if(ans[last] == n)
return ;
return ;
}
for(int i=last; i>=; i--)
{
for(int j=i; j>=; j--)
{
int tmp = ans[i] + ans[j];
if(tmp <= n && !vis[tmp])
{
if(tmp < ans[last])
return ;
vis[tmp] = ;
ans[last+] = tmp;
if(dfs(last+,lim))
return ;
}
}
}
return ;
} int main()
{
while(~scanf("%d",&n) && n)
{
ans[] = ;
for(int i=; i<=; i++)
{
if(dfs(,i))
{
for(int j=; j<=i; j++)
{
printf(j == i?"%d\n":"%d ",ans[j]);
}
break;
}
}
}
}
dfs剪枝：
#include<iostream>
#include<cstdio>
#include<string.h>
using namespace std; int ans[];
int t_ans[];
int b[];
bool vis[];
int n,m; void init()
{
m = n;
b[n] = ;
t_ans[] = ;
for(int i=n-;i>=;i--)
{
b[i] = b[min(i+i,n)] + ;
}
} void dfs(int last)
{
bool vis[];
memset(vis,,sizeof(vis));
if(last + b[t_ans[last]] > m)return;
if(t_ans[last] == n)
{
if(last <= m)
{
m = last;
for(int i=;i<=last;i++)
{
ans[i] = t_ans[i];
}
}
return;
}
for(int i=last; i>=; i--)
{
for(int j=i; j>=; j--)
{
int tmp = t_ans[i] + t_ans[j];
if(tmp <= n && !vis[tmp])
{
if(tmp < t_ans[last])
return;
vis[tmp] = ;
t_ans[last+] = tmp;
dfs(last+);
}
}
}
return;
} int main()
{
while(~scanf("%d",&n) && n)
{
init();
dfs();
for(int i=;i<=m;i++)
{
printf(i == m?"%d\n":"%d ",ans[i]);
}
}
}