2541 幂运算

2541 幂运算

 

 时间限制: 1 s

 空间限制: 128000 KB

 题目等级 : 钻石 Diamond

题解

 查看运行结果

 

 

题目描述 Description

    从m开始，我们只需要6次运算就可以计算出m31：

 

    m2=m×m，m4=m2×m2，m8=m4×m4，m16=m8×m8，m32=m16×m16，m31=m32÷m。

 

    请你找出从m开始，计算mn的最少运算次数。在运算的每一步，都应该是m的正整数次方，换句话说，类似m-3是不允许出现的。

输入描述 Input Description

输入为一个正整数n

输出描述 Output Description

输出为一个整数，为从m开始，计算mn的最少运算次数。

样例输入 Sample Input

样例1
1

样例2
31

样例3
70

样例输出 Sample Output

样例1
0

样例2
6

样例3
8

数据范围及提示 Data Size & Hint

n（1<=n<=1000）

 

数据没有问题，已经出现过的n次方可以直接调用

分类标签 Tags 点此展开 

 

深度优先搜索 搜索

史上最有潜力的打表，快来围观！

#include<iostream>
using namespace std;
int a[1001]={0,0,1,2,2,3,3,4,3,4,4,5,4,5,5,5,4,5,5,6,5,
6,6,6,5,6,6,6,6,7,6,6,5,6,6,7,6,7,7,7,6,
7,7,7,7,7,7,7,6,7,7,7,7,8,7,8,7,8,8,8,7,
8,7,7,6,7,7,8,7,8,8,8,7,8,8,8,8,8,8,8,7,
8,8,8,8,8,8,9,8,9,8,9,8,8,8,8,7,8,8,8,8,
9,8,9,8,9,9,9,8,9,9,9,8,9,9,9,9,9,9,9,8,
9,9,9,8,9,8,8,7,8,8,9,8,9,9,9,8,9,9,9,9,
9,9,9,8,9,9,9,9,9,9,10,9,9,9,9,9,9,9,9,8,
9,9,9,9,9,9,10,9,10,9,10,9,10,10,10,9,10,10,10,9,
10,10,10,9,10,9,10,9,9,9,9,8,9,9,9,9,10,9,10,9,
10,10,10,9,10,10,10,9,10,10,10,10,10,10,10,9,10,10,10,10,
10,10,10,9,10,10,10,10,10,10,10,10,10,10,10,10,10,10,10,9,
10,10,10,10,10,10,10,9,10,10,10,9,10,9,9,8,9,9,10,9,
10,10,10,9,10,10,11,10,11,10,10,9,10,10,11,10,11,10,10,10,
10,10,10,10,10,10,10,9,10,10,10,10,10,10,11,10,10,10,11,10,
11,11,11,10,11,10,11,10,11,10,11,10,11,10,10,10,10,10,10,9,
10,10,10,10,10,10,11,10,11,10,11,10,11,11,11,10,11,11,11,10,
11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,
11,11,11,11,11,11,11,10,11,11,11,10,11,11,11,10,11,10,11,10,
10,10,10,9,10,10,10,10,11,10,11,10,11,11,11,10,11,11,11,10,
11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,10,11,11,11,11,
11,11,11,11,11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,
11,11,11,11,11,11,11,10,11,11,11,11,11,11,11,11,11,11,11,11,
12,11,12,11,11,11,12,11,12,11,11,11,11,11,11,11,11,11,11,10,
11,11,11,11,11,11,11,11,11,11,12,11,12,11,11,10,11,11,12,11,
12,11,11,10,11,11,11,10,11,10,10,9,10,10,11,10,11,11,11,10,
11,11,12,11,12,11,11,10,11,11,12,11,12,12,11,11,12,12,12,11,
12,11,11,10,11,11,12,11,12,12,12,11,11,12,12,11,12,11,12,11,
11,11,12,11,12,11,11,11,12,11,11,11,11,11,11,10,11,11,11,11,
11,11,12,11,11,11,12,11,12,12,12,11,12,11,12,11,12,12,12,11,
12,12,12,12,12,12,12,11,12,12,12,11,12,12,12,11,12,12,12,11,
12,12,12,11,12,12,12,11,12,11,12,11,12,11,11,11,11,11,11,10,
11,11,11,11,11,11,12,11,12,11,12,11,12,12,12,11,12,12,12,11,
12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,
12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,
12,12,12,11,12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,11,
12,12,12,12,12,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,
12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,11,12,12,12,11,
12,11,12,11,11,11,11,10,11,11,11,11,12,11,12,11,12,12,12,11,
12,12,12,11,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,11,
12,12,12,12,12,12,12,12,12,12,13,12,12,12,12,11,12,12,12,12,
13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,12,12,12,12,
12,12,12,12,12,12,13,12,12,12,13,12,13,12,12,12,13,12,12,12,
12,12,12,11,12,12,12,12,12,12,13,12,12,12,13,12,13,12,12,12,
13,12,13,12,13,12,12,12,12,12,12,12,12,12,12,11,12,12,12,12,
12,12,12,12,12,12,13,12,13,12,13,12,13,12,13,12,13,12,13,12,
13,13,13,12,13,13,13,12,13,12,13,12,13,13,13,12,13,13,13,12,
13,12,13,12,12,12,13,12,13,12,12,12,12,12,12,12,12,12,12,11,
12,12,12,12,12,12,12,12,12,12,13,12,13,12,12,12,12,12,13,12,
13,13,13,12,13,13,13,12,13,12,12,11,12,12,13,12,13,13,13,12
};
int main(){
    int n;
    cin>>n;
    cout<<a[n]<<endl;
    return 0;
}

 

正解：仿照 快速幂（上面也AC了）

本代码做不到 0ms

#include<iostream>
using namespace std;
#include<cstdio>
#define MAXdeep 20
int a[MAXdeep];
bool dfs(int k,int maxdepth,int n)
{
    if(a[k]==n) return true;
    if(k==maxdepth) return false;
    int maxx=a[0];
    for(int i=1;i<=k;++i)
      maxx=max(maxx,a[i]);
    if((maxx<<(maxdepth-k))<n) return false;
    for(int i=k;i>=0;--i)//
    {
        a[k+1]=a[i]+a[k];
        if(dfs(k+1,maxdepth,n)) return true;
        a[k+1]=a[k]-a[i];
        if(dfs(k+1,maxdepth,n)) return true;
    }
    return false;
}
int solve(int n)
{
    if(n==1) return 0;
    a[0]=1;
    for(int i=1;i<MAXdeep;++i)
      if(dfs(0,i,n)) return i;
    return MAXdeep;
}
int main()
{
    int n;
    scanf("%d",&n);
    printf("%d\n",solve(n));
    return 0;
}