POJ 2248

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

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

#include <iostream>
#include <fstream>
#include <sstream>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <set>
#include <map>
#include <list>
#include <iomanip>
#include <cctype>
#include <cassert>
#include <bitset>
#include <ctime>

using namespace std;

#define pau system("pause")
#define ll long long
#define pii pair<int, int>
#define pb push_back
#define pli pair<ll, int>
#define pil pair<int, ll>
#define clr(a, x) memset(a, x, sizeof(a))

const double pi = acos(-1.0);
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double EPS = 1e-9;

/*
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
#define TREE tree<pli, null_type, greater<pli>, rb_tree_tag, tree_order_statistics_node_update>
TREE T;
*/


int n, num[27];
bool dfs(int dep, int k) {
    if (n == num[k]) {
        for (int i = 1; i < k; ++i) {
            printf("%d ", num[i]);
        }
        printf("%d
", num[k]);
        return true;
    }
    for (int j = 1; j <= k; ++j) {
        num[k + 1] = num[k] + num[j];
        if (num[k + 1] > n || k + 1 > dep) continue;
        if (dfs(dep, k + 1)) return true;
    }
    return false;
}
int main() {
    num[1] = 1;
    while (scanf("%d", &n) && n) {
        for (int dep = 1; ; ++dep) {
            if (dfs(dep, 1)) break;
        }
    }
    return 0;
}

解法：迭代加深搜索，扩展可等加成每次由最后一个和前面的相加。