Project Euler：Problem 32 Pandigital products-白红宇

Project Euler：Problem 32 Pandigital products

阅读量：5949 次

发布时间：2019-06-19

本文共 2013 字，大约阅读时间需要 6 分钟。

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

We shall say that an n-digit number is pandigital if it makes use of all the digits 1 to n exactly once; for example, the 5-digit number, 15234, is 1 through 5 pandigital.

The product 7254 is unusual, as the identity, 39 × 186 = 7254, containing multiplicand, multiplier, and product is 1 through 9 pandigital.

Find the sum of all products whose multiplicand/multiplier/product identity can be written as a 1 through 9 pandigital.

HINT: Some products can be obtained in more than one way so be sure to only include it once in your sum.

#include 
     
      #include 
      using namespace std;bool pand(int a, int b, int c){	map
       
        mp;	int tmp[] = { a, b, c };	int num = 1;	if (b != 0)		num = 3;	for (int i = 0; i < num; i++)	{		int p = tmp[i];		while (p)		{			if (mp[p % 10] != 0)				return false;			else			{				mp[p % 10] = 1;				p = p / 10;			}		}	}	if (mp[0] != 0)		return false;	if (b == 0)         //说明a中无反复		return true;	int count = 0;	for (int i = 1; i <= 9; i++)	{		if (mp[i] != 0)			count++;	}	if (count == 9)		return true;	else		return false;}int main(){	map
        
         mp;	for (int i = 1; i <= 9876; i++)	{		if (pand(i,0,0))		{			for (int j = 1; j < i; j++)			{				if (i*j <= 10000)				{					if (pand(i, j, i*j))						mp[i*j] = 1;				}			}		}	}	map
         
          ::iterator iter;	int res = 0;	for (iter = mp.begin(); iter != mp.end(); iter++)	{		if (mp[iter->first] == 1)			res += iter->first;	}	cout << res << endl;	system("pause");	return 0;}

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