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 <iostream>
#include <map>
using namespace std;
bool pand(int a, int b, int c)
{
map<int, int>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中无重复
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<int, int>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<int, int>::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;
}