POJ 2248 Addition Chains

Description

An addition chain for n is an integer sequence 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.
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

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由于数据不大，可以枚举数列长度，迭代加深搜索。
由于数列严格单调递增，所以可以记录末尾值剪枝。

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#include
#include
using namespace std;
int n,dep,a[101];
bool dfs(int cur,int lst)
{
    if(cur==dep)
    {
        if(a[cur-1]==n)
            return 1;
        return 0;
    }
    for(int i=0;i<=cur-1;i++)
        for(int j=i;j<=cur-1;j++)
        {
            if(a[i]+a[j]>n)
                break;
            if(a[i]+a[j]>lst)
            {
                a[cur]=a[i]+a[j];
                if(dfs(cur+1,a[cur]))
                    return 1;
            }
        }
    return 0;
}
int main()
{
    a[0]=1;
    while(scanf("%d",&n)&&n)
    {
        for(int i=1;;i++)
        {
            dep=i;
            if(dfs(1,1))
            {
                for(int j=0;j<=i-1;j++)
                    printf("%d ",a[j]);
                printf("\n");
                break;
            }
        }
    }
    return 0;
}