Problem 44 Pentagon numbers （暴力...）

古城微笑少年丶 2022-07-15 02:43 80阅读 0赞

## Pentagon numbers ##

### Problem 44 ###

Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2. The first ten pentagonal numbers are:

1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...

It can be seen that P4 \+ P7 = 22 + 70 = 92 = P8. However, their difference, 70 − 22 = 48, is not pentagonal.

Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?

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         <span style="font-size:18px">Answer:</span> 
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       <td><strong><span style="font-size:18px">5482660</span></strong></td> 
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       <td colspan="2"><span style="color:rgb(153,153,153)"><span style="font-size:18px">Completed on Sun, 30 Oct 2016, 08:19</span></span></td> 
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题解：无脑暴力...

代码：

#include<bits/stdc++.h>
    using namespace std;
    int calc(int n)
    {
    	return n*(3 * n - 1) / 2;
    }
    int solve(int n)
    {
    	for (int i = 1; i*(3*i-1) <= 2*n; i++)
    	{
    		if (i*(3 * i - 1) == 2 * n)
    			return 1;
    	}
    	return 0;
    }
    int main()
    {
    	int ans=0;
    	for (int i = 2; i <= 3000; i++)
    	{
    		for (int j = 1; j < i; j++)
    		{
    			int sum = calc(i) + calc(j);
    			int diff = calc(i) - calc(j);
    			if (solve(sum) && solve(diff))
    				cout << diff << endl;
    		}
    	}
    	return 0;
    }

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