暴力求解法，由於數據規模小，窮舉即可，注意取位的技巧。
代碼如下：
/*
ID: michael139
LANG: C
PROG: crypt1
*/
#include<stdio.h>
#include<string.h>
int vis[10];
int main () {
FILE *fin = fopen("crypt1.in", "r");
FILE *fout = fopen("crypt1.out", "w");
int n,i,temp,abc,de,min,max,x,y,z,p,q,mid1,mid2,ans,count;
while (fscanf(fin,"%d",&n) != EOF) {
count = 0;
memset(vis,0,sizeof(vis));
min = 10;
max = -1;
for (i=1;i<=n;i++) {
fscanf(fin,"%d",&temp);
vis[temp] = 1;
if (min>temp) min = temp;
if (max<temp) max = temp;
}
for (abc=100*min+10*min+min;abc<=100*max+10*max+max;abc++) {
for (de=10*min+min;de<=10*max+max;de++) {
x = abc/100;
y = (abc - x*100)/10;
z = abc%10;
if (!vis[x] || !vis[y] || !vis[z]) break;
p = de/10;
q = de%10;
if (!vis[p] || !vis[q]) continue;
mid1 = q*abc;
mid2 = p*abc;
if (mid1+mid2*10>=10000 || mid1+mid2*10<1111) continue;
x = mid1/100;
y = (mid1 - x*100)/10;
z = mid1%10;
if (!vis[x] || !vis[y] || !vis[z]) continue;
x = mid2/100;
y = (mid2 - x*100)/10;
z = mid2%10;
if (!vis[x] || !vis[y] || !vis[z]) continue;
ans = mid1+mid2*10;
x = ans/1000;
y = (ans - x*1000)/100;
z = (ans - x*1000 - y*100)/10;
p = ans%10;
if (vis[x] && vis[y] && vis[z] && vis[p]) count++;
}
}
fprintf(fout,"%d\n",count);
}
return 0;
}