uva-529-枚举

题意:

a0,a1,a2,a3....an

对于任意的i,j,k

0<=k<=n

0<=i<=k-1

0<=j<=k-1

ak=ai+aj

求a0.....an

解题思路:

迭代加深搜索,限定下界的深搜

首页,利用贪心算法,求出最小的步数depth,an=2ai,当an>=n的时候即可得到最小元素个数,由当前步数一步一步加大搜索的步数.

#include <stdio.h>
#include<iostream>
#include <string.h>
#include<memory.h>
using namespace std;
const int N = ;
int n = ;
int ans[N] = {  };
int depth = ;
int flag = ;
void dfs(int cur)
{
    if(flag)
        return;
    if(cur == depth)
    {
        if(ans[cur] == n)
            flag = ;
        return;
    }
    for(int i = ; i <= cur; i++)
    {
        for(int j = i; j <= cur; j++)
        {
            if(ans[i] + ans[j] > ans[cur] && ans[i] + ans[j] <= n)
            {
                int sum = ans[i] + ans[j];
                for(int k = cur + ; k <= depth; k++)
                    sum = sum * ;
                if(sum < n)
                    continue;
                ans[cur + ] = ans[i] + ans[j];
                dfs(cur + );
                if(flag)
                    return;
            }
        }

    }
}
int main()
{
    freopen("d://1.txt", "r", stdin);
    while (cin >> n && n)
    {
        memset(ans, , sizeof(ans));
        ans[] = ;
        int k = ;
        depth = ;
        flag = ;
        while (k < n)
        {
            ++depth;
            k = k * ;
        }
        while (!flag)
        {
            dfs();
            if(!flag)
                ++depth;
        }
        cout << ans[];
        for(int i = ; i <= depth; i++)
            cout << " " << ans[i];
        cout << endl;
    }
    return ;
}