Prime Cryptarithm
The following cryptarithm is a multiplication problem that can be
solved by substituting digits from a specified set of N digits into the
positions marked with *. If the set of prime digits {2,3,5,7} is
selected, the cryptarithm is called a PRIME CRYPTARITHM.
* * *
x * *
-------
* * * <-- partial product 1
* * * <-- partial product 2
-------
* * * *
Digits can appear only in places marked by `*'. Of course, leading
zeroes are not allowed.
Note that the 'partial products' are as taught in USA schools.
The first partial product is the product of the final digit of the
second number and the top number. The second partial product is
the product of the first digit of the second number and the top
number.
Write a program that will find all solutions to the cryptarithm
above for any subset of digits from the set {1,2,3,4,5,6,7,8,9}.
PROGRAM NAME: crypt1
INPUT FORMAT
| Line 1: |
N, the number of digits that will be used |
| Line 2: |
N space separated digits with which to solve the cryptarithm |
SAMPLE INPUT (file crypt1.in)
5
2 3 4 6 8
OUTPUT FORMAT
A single line with the total number of unique solutions. Here
is the single solution for the sample input:
2 2 2
x 2 2
------
4 4 4
4 4 4
---------
4 8 8 4
SAMPLE OUTPUT (file crypt1.out)
1
方法是暴利枚舉然后判斷 - -！
/*
ID: superlo1
LANG: C++
TASK: crypt1
*/
#include <iostream>
using namespace std;
int n, set[10];
bool hash[11];
bool check(int x, int flag)
{
int a[4], len = 0;
while(x)
{
a[++len] = x % 10;
x = x / 10;
if(hash[a[len]] == 0) return false;
}
if(len > 3 && !flag) return false;
return true;
}
int main()
{
freopen("crypt1.in","r",stdin);
freopen("crypt1.out","w",stdout);
scanf("%d", &n);
for(int i = 0; i < n; i ++)
{
scanf("%d", &set[i]);
hash[set[i]] = 1;
}
int i, j, cnt = 0;
for(i = 100; i <= 999; i ++)
if(check(i, 0))
{
for(j = 10; j <= 99; j ++)
if(check(j, 0) && check( (j % 10) * i, 0)
&& check( (j / 10) * i, 0) && check(i * j, 1))
{
//printf("%d %d\n",i,j);
cnt ++;
}
}
printf("%d\n",cnt);
//while(1);
}