Codeforces 342A Xenia and divisors (water problem)

Test instructions: given n number (less than or equal to 7), let you divide it into M-groups, each group has three numbers, and satisfies, a < b < C, and a can be divisible by b,b divisible by C.

Analysis: For this problem, because the topic said is not more than 7, then a think not on the three groups, 124,136,126. In the three groups, and then determine the number of each group, first only the second group has 3, then the number of the second group is determined,

Then look at the remaining two groups, only the first group has 4, then the first group is determined, then the rest is the third group, of course, the third group only 6.

The code is as follows:

#include <bits/stdc++.h>using namespace Std;const int maxn = 33333 + 5;typedef long Long ll;int a[10];int main () {
   int n, x;    while (CIN >> N) {        memset (a, 0, sizeof (a));        for (int i = 0; i < n; ++i) {            cin >> x;            ++A[X];        }        int m = N/3;        bool OK = true;        if (A[5] | | a[7]) OK = false;        if (A[1] < a[3] | | a[6] < a[3]) OK = false;        A[1]-= a[3], a[6]-= a[3];        int a1 = a[3];        if (a[1]! = a[2] | | a[6] + a[4]! = a[2]) ok = false;        int a2 = a[6];        int a4 = a[4];        if (!ok)  printf (" -1\n");        else{for            (int i = 0; i < A1; ++i)  printf ("1 3 6\n");            for (int i = 0; i < A2; ++i)  printf ("1 2 6\n");            for (int i = 0; i < A4; ++i)  printf ("1 2 4\n");        }    }    return 0;}

Codeforces 342A Xenia and divisors (water problem)