1.4.3 USACO Prime Cryptarithm（枚舉）

Prime Cryptarithm

(This poorly named task has nothing to do with prime numbers or even, really, prime digits. Sorry 'bout that.)

A cryptarithm is usually presented as a pencil-and-paper task in which the solver is required to substitute a digit for each of the asterisks (or, often, letters) in the manual evaluation of an arithmetic term or expression so that the consistent application of the digits results in a proper expression. A classic example is this cryptarithm, shown with its unique solution:

    SEND            9567       S->9  E->5  N->6  D->7
  + MORE          + 1085       M->1  O->0  R->8
  -------        -------
   MONEY           10652       Y->2

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 *. Since the asterisks are generic, any digit from the input set can be used for any of the asterisks; any digit may be duplicated as many times as desired.

Consider using the set {2,3,5,7} for the cryptarithm below:

      * * *
   x    * *
    -------
      * * *         <-- partial product 1 -- MUST BE 3 DIGITS LONG
    * * *           <-- partial product 2 -- MUST BE 3 DIGITS LONG
    -------
    * * * *

Digits can appear only in places marked by `*'. Of course, leading zeroes are not allowed.

The partial products must be three digits long, even though the general case (see below) might have four digit partial products. 

********** Note About Cryptarithm's Multiplication ************ 
In USA, children are taught to perform multidigit multiplication as described here. Consider multiplying a three digit number whose digits are 'a', 'b', and 'c' by a two digit number whose digits are 'd' and 'e':

[Note that this diagram shows far more digits in its results than
the required diagram above which has three digit partial products!]

          a b c     <-- number 'abc'
        x   d e     <-- number 'de'; the 'x' means 'multiply'
     -----------
p1      * * * *     <-- product of e * abc; first star might be 0 (absent)
p2    * * * *       <-- product of d * abc; first star might be 0 (absent)
     -----------
      * * * * *     <-- sum of p1 and p2 (e*abc + 10*d*abc) == de*abc

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 supplied non-zero single-digits. Note that the multiplicands, partial products, and answers must all conform to the cryptarithm's framework.

PROGRAM NAME: crypt1

INPUT FORMAT

Line 1:	N, the number of digits that will be used
Line 2:	N space separated non-zero 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 solutions. Here is the one and only 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

OUTPUT DETAILS

Here's why 222x22 works: 3 digits times 2 digits yields two (equal!) partial products, each of three digits (as required). The answer has four digits, as required. Each digit used {2, 4, 8} is in the supplied set {2, 3, 4, 6, 8}.

Why 222x23 doesn't work:

      2 2 2   <-- OK:  three digits, all members of {2, 3, 4, 6, 8}
        2 3   <-- OK:  two digits, all members of {2, 3, 4, 6, 8}
     ------
      6 6 6   <-- OK:  three digits, all members of {2, 3, 4, 6, 8}
    4 4 4     <-- OK:  three digits, all members of {2, 3, 4, 6, 8}
  ---------
    5 1 0 6   <-- NOT OK: four digits (good), but 5, 1, and 0 are not in
                                                    {2, 3, 4, 6, 8}
 

Submit a solution:

翻譯：P1211 [USACO1.3]牛式 Prime Cryptarithm

思路：其實本題就是求一個三位數x和兩位數y相乘得到結果sum，判斷x,y,sum以及x*(y%10),x*(y/10)的每一位數是否出現在給定的數字中即可。

這裏只需要枚舉每一個三位數和兩位數相乘，判斷是否爲牛式即可。

 

/*
ID: L
PROG: crypt1
LANG: C++
*/
#include<bits/stdc++.h> 
using namespace std;
bool a[11];
bool ex(int k)//判斷k中的每位數是否在a中 
{
	while(k)
	{
		if(a[k % 10] == 0) return 0;
		k = k /10;
	}
	return 1;
}
bool check(int x, int y)//是否滿足牛式，x代表三位數，y代表2位數 
{
	int y1 = y%10;//分離y的個位數 
	int y2 = y/10;//分離y的十位數 
	int p = x*y1, q = x*y2;
	int sum = x*y;
	if(p > 999 || q >999 || sum > 9999) return 0;
	if(ex(x)==1 && ex(y)==1 && ex(p) == 1 && ex(q) == 1 && ex(sum) == 1) return 1; //判斷數字是否是測試數據中的數字 
	else return 0;
}
int main()
{
	freopen("crypt1.in","r",stdin);
	freopen("crypt1.out","w",stdout);
	int n,m,ans=0;
	cin >> n;
	for(int i = 1; i <= n; ++i)
	{
		cin >> m;
		a[m] = 1;//出現過 
	}
	for(int i = 100; i <= 999; ++i)//枚舉三位數與兩位數的組合 
	{
		for(int j = 10; j <= 99; ++j)
		{
			if(check(i,j))  ans++;
		} 
	} 
	cout << ans << endl;
	fclose(stdin);
	fclose(stdout);
	return 0;
}