UVA 529 Addition Chains

分析

给定一个数，尝试构造一个最短的数列，每一个数字由其之前的数字相加得，使得最后一个数字为所给数字。

每一次尝试相加其实应该是前一个值尝试加上自它向前的所有值。例如5，尝试1，2，…，下一位应该先尝试4，再尝试3。

这里注意需要剪枝！如何判断能否构造成功呢？
考虑一下上面的尝试方法，最大值是不是它的2倍呢？如果这个值在有限次数的翻倍也不能到达这个数，那么可以不用尝试了。

至于最大长度16个数，还是一个magic number（提交多次试的..当然无妨开大一些的)。

代码

#include <cstdio>#define MAX_N 16int n, p, a[MAX_N] = {1}, r[MAX_N];void dfs(int t){    if (t >= p) return;    if (a[t-1] == n) {        p = t;        for (int i = 0; i < p; i++) r[i] = a[i];        return;    }    for (int i = t-1; i >= 0; i--) {        a[t] = a[t-1] + a[i];        if (a[t] <= n && a[t]<<(p-t-1) >= n) dfs(t + 1);    }}void solve(){    p = MAX_N;    dfs(1);    for (int i = 0; i < p; i++) {        printf("%d", r[i]);        if (i != p-1) printf(" ");        else printf("\n");    }}int main(){    while (scanf("%d", &n), n) solve();}

题目

Description

An addition chain for n is an integer sequence <a0,a1,a2,…,am> with the following four properties:

• a₀ = 1
• a_m = n
• a₀< a₁< a₂<…< a_m-1< a _m
• For each k ( 1≤k≤m) there exist two (not neccessarily different) integers i and j ( 0≤i,j≤k−1) with a_k=a_i+a_j

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 Specification

The input file will contain one or more test cases. Each test case consists of one line containing one integer n ( 1≤n≤10000). Input is terminated by a value of zero (0) for n.

Output Specification

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

 571215770

Sample Output

1 2 4 51 2 4 6 71 2 4 8 121 2 4 5 10 151 2 4 8 9 17 34 68 77