ZOJ Problem Set 2952 Find All M^N Please

Find All M^N Please

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Time Limit: 1 Second      Memory Limit: 32768 KB

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Recently, Joey has special interest in the positive numbers that could be represented as M ^ N (M to the power N), where M and N are both positive integers greater than or equal to 2. For example, 4, 8, 9 and 16 are first four such numbers, as 4 = 2 ^ 2, 8 = 2 ^ 3, 9 = 3 ^ 2, 16 = 2 ^ 4. You are planning to give Joey a surprise by giving him all such numbers less than 2 ^ 31 (2147483648). List them in ascending order, one per line.

Sample Output

4
8
9
16
25
27
32
|
| <-- a lot more numbers
|
1024
1089
1156
1225
1296
1331
|
|
|

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Author: SHEN, Guanghao
Source: Zhejiang University Local Contest 2008

SOURCE CODE :

#include<iostream>
#include<map>
using namespace std;

int main()
{
    map<int, int> ma;
    for(int M = 2; M <= 46340;M++)
    {
        long long int pow = M;
        if(ma.find(M) != ma.end()) continue;
        for(int N = 1; N < 32; N++)
        {
            pow *= M;
            if(pow >= 2147483648) break;
            ma[pow] = pow;
        }
    }
    for(map<int,int>::iterator it = ma.begin();it != ma.end(); it++)
    {
        cout<<it->first<<endl;
    }
    return 0;
}

在第二层for循环的时候，我曾想过先算出最大的可能循环数，然后在进行循环，这样就省去了

if(pow >= 2147483648) break;

的判断，但是怎么样才能判断最大循环次数呢？我用了log(numeric_limits<int>::max() * 1.0)/log(M*1.0)来计算，本来我还以为这样计算是最大的节省了时间，但是我错了，cmath的log方法，算法肯定不简单，因为我发现，就算我完成所有的计算，这个循环次数的耗时，占了总耗时的差不多20%左右的时间开销。看来，有时候还是不要自作聪明的好。