POJ 2248 Addition Chains

Addition Chains

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 4480		Accepted: 2454		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

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
题目大意：输入一个数字n,输出一个数列,要求数列的第一个为1，后面的由前面的数字是前面数字中人选两个想家的和，可以选两个相同的数字，亦即对于任意i和j (0<=i, j<=k-1) 有 ak=ai+aj，输出长度最短的一个序列。

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;

int MinLen;
int n;
int result[10005] = {1, 2};
int temp[10005] = {1, 2};

void DFS(int nLen, int num)
{
    if (nLen >= MinLen || num > n)
    {
        return;
    }
    if (num == n)
    {
        if (MinLen > nLen)
        {
            MinLen = nLen;
            for (int i = 2; i < nLen; i++)
            {
                result[i] = temp[i];
            }
        }
        return;
    }
    for (int i = nLen - 1; i >= 0; i--)
    {
        temp[nLen] = temp[i] + temp[nLen - 1];
        DFS(nLen + 1, temp[nLen]);
    }
}

int main()
{
    while(scanf("%d", &n) != EOF && n != 0)
    {
        if (n == 1)
        {
            printf("1\n");
            continue;
        }
        MinLen = 0x7fffffff;
        DFS(2, 2);
        printf("%d", result[0]);
        for (int i = 1; i < MinLen; i++)
        {
            printf(" %d", result[i]);
        }
        printf("\n");
    }
    return 0;
}