Addition Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 5454		Accepted: 2923		Special Judge

Description

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 < ... < a_m-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

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

Source

Ulm Local 1997

Submission address: poj

Search for a bit of k in turn, enumerate the previous i, j, add a[i] + a[j] to the position of a[k], and then continue to search;

Pruning: try to make the number of sequences approach n as soon as possible from reaching a small enumeration i, j;

In order not to repeat the search, use a bool array to store whether a[i] + a[j] has been searched;

Then because the depth of the answer is very small, it is iteratively deepened in one shot;

This is almost A;

Here is the code:

//By zZhBr
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int n;
int years;

int a[1100];

bool use[1005];
bool DFS(int stp)
{
    memset(use, 0, sizeof use);
    
    if(stp > ans)
    {
        if(a[ans] == n) return 1;
        else return 0;
    }
    
    for(register int i = stp - 1 ; i >= 1 ; i --)
    {
        for(register int j = i ; j >= 1 ; j --)
        {
            if(a[i] + a[j] > n) continue;
            if(!use[a[i] + a[j]])
            {
                if(a[i] + a[j] <= a[stp - 1]) return 0;
                use[a[i] + a[j]] = 1;
                a[stp] = a[i] + a[j];
                if(DFS(stp + 1)) return 1;
                a[stp] = 0;
                use[a[i] + a[j]] = 0;
            }
        }
    }
}

intmain ()
{
    while(scanf("%d", &n) != EOF)
    {
        if(n == 0) return 0;
        if(n == 1)
        {
            printf("1\n");
            continue;
        }
        if(n == 2)
        {
            printf("1 2\n");
            continue;
        }
        a[1] = 1;a[2] = 2;
        for(ans = 3 ; !DFS(3) ; ans ++);
        for(register int i = 1 ; i <= ans ; i ++)
        {
            printf("%d ", a[i]);
        }
        printf("\n");
        memset(a, 0, sizeof a);
    }
    return 0;
}

zZhBr

poj2248 Addition Chains iteratively deepen the search