USACO section 1.3.4 Prime Cryptarithm

/*
ID: dollar4
PROG: crypt1
LANG: C++
*/
#include <iostream>
#include <fstream>
#include <string>
#include <algorithm>
#include <cstring>

using namespace std;
int num[10], n;
bool cmp(int a, int b)
{
return a < b;
}
bool findin(int a)
{
for (int i = 0; i < n; i++)
if (num[i] == a)
return true;
return false;
}

int main()
{
ofstream fout ("crypt1.out");
ifstream fin ("crypt1.in");
fin >> n;
int i, j, k, l, m, s1, s2, s3, s4, s5;
int cnt = 0, mm, nn, mn;
for (i = 0; i < n; i++)
fin >> num[i];
sort(num, num + n, cmp);
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
for (k = 0; k < n; k++)
for (l = 0; l < n; l++)
for (m = 0; m < n; m++)
{
s1 = num[i];
s2 = num[j];
s3 = num[k];
s4 = num[l];
s5 = num[m];
mm = s4 * (s1 * 100 + s2 * 10 + s3);
nn = s5 * (s1 * 100 + s2 * 10 + s3);
mn = nn + 10 * mm;
if (s1 * s4 > 9 || s5 * s1 > 9)//相乘必须是3位数
continue;
//想加的约束
if (s1 * s4 + s2 * s4 / 10 >= 10 || s1 * s5 + s2 * s5 / 10 >= 10)//nn / 100 + mm % 100 / 10 < 10 ||
continue;
//运算得到的结果有没有超出所给的数组之外的
if ( !findin(nn / 100) || !findin(nn % 100 / 10) || !findin(nn % 10) || !findin(mm / 100) || !findin(mm % 100 / 10) || !findin(mm % 10))
continue;
if (!findin(mn / 1000) || !findin(mn % 1000 / 100) || !findin(mn % 100 / 10) || !findin(mn % 10))
continue;
cnt++;
}
fout << cnt << endl;
return 0;
}

2. 官方代码：

The constraints of this problem are small enough that we can just try all possible products of 3 digit * 2 digit numbers, and look to see if all the correct digits are used.The function "isgood" checks that a number is composed only of acceptable digits, and "isgoodprod" checks that all the lines of the multiplication are composed of acceptable digits.#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>

int isgooddigit[10];	/* isgooddigit[d] is set if d is an acceptable digit
*/

/* check that every decimal digit in "n" is a good digit,
and that it has the right number "d" of digits. */
int
isgood(int n, int d)
{
if(n == 0)
return 0;

while(n) {
if(!isgooddigit[n%10])
return 0;
n /= 10;
d--;
}

if( d == 0 )
return 1;
else
return 0;
}

/* check that every product line in n * m is an okay number */
int
isgoodprod(int n, int m)
{
if(!isgood(n,3) || !isgood(m,2) || !isgood(n*m,4))
return 0;

while(m) {
if(!isgood(n*(m%10),3))
return 0;
m /= 10;
}
return 1;
}

void
main(void)
{
int i, j, n, nfound;
FILE *fin, *fout;

fin = fopen("crypt1.in", "r");
fout = fopen("crypt1.out", "w");
assert(fin != NULL && fout != NULL);

for(i=0; i<10; i++) {
isgooddigit[i] = 0;
}
fscanf(fin, "%d", &n);
for(i=0; i<n; i++) {
fscanf(fin, "%d", &j);
isgooddigit[j] = 1;
}

nfound = 0;
for(i=100; i<1000; i++)
for(j=10; j<100; j++)
if(isgoodprod(i, j))
nfound++;

fprintf(fout, "%d\n", nfound);
exit(0);
}