OJ in class: Determine the password strength and oj password strength.

OJ in class: Determine the password strength and oj password strength.

One problem in determining password strength:

Assume that the following four types of characters are allowed as passwords:
(1) uppercase English letters, (2) lowercase English letters, (3) numbers 0-9, (4) special symbols @-_#~
Set the password strength as follows:
Best: The length is greater than or equal to 16 and contains the above four types of characters. Each type has at least two different characters.
Strong: does not comply with the Best rules, but the length is greater than or equal to 10, and contains the above four types of Characters
Medium: does not comply with the Best and Strong rules, but the length is greater than or equal to 8, and contains at least three types of Characters
Weak: does not comply with Best, Strong, and Medium rules.
Here is a password combination. Determine the password strength.
Input Format: a line of string (length <100, ending with a line break), and must be of the above four types of characters.
Output Format: one line of information. The length of the password is output first, followed by the strength level, with a blank space in the middle.

Input Example 1: abcdXYZ1234 @-_-#
Output example Best
Input Example 2: 12345678 abcdefghXYZ
Sample output: Medium
Input Example 3: aX1 @
Output Example 3: 4 Weak

 

The key to the question is that each type of Best has at least two different characters. To implement this function, you can define multiple arrays.

C source code: # include "stdio. h"
Int main ()
{
Char a [100], sz [100], ZM [100], zm [100], fh [100];
Int I, j, k, l, m, n, sum, zm0, ZM0, sz0, fh0;
I = 0;
K = l = m = n = 0;
Zm0 = 0;
ZM0 = 0;
Sz0 = 0;
Fh0 = 0;
Sum = 0;
While (a [I] = getchar ())! = '\ N ')
{
I ++;
}
For (j = 0; j <I; j ++)
{
If (a [j]> = 'A' & A [j] <= 'Z ')
{
ZM [k] = a [j];
K ++;
}

Else if (a [j]> = 'A' & a [j] <= 'Z ')
{
Zm [l] = a [j];
L ++;
}
Else if (a [j]> = '0' & a [j] <= '9 ')
{
Sz [m] = a [j];
M ++;
}
Else
{
Fh [n] = a [j];
N ++;
}
}
Sum = k + l + m + n;
While (k --)
{
ZM0 ++;
If (ZM [k] = ZM [k-1])
ZM0 --;
}
While (l --)
{
Zm0 ++;
If (zm [l] = zm [L-1])
Zm0 --;
}
While (m --)
{
Sz0 ++;
If (sz [m] = sz M-1])
Sz0 --;
}
While (n --)
{
Fh0 ++;
If (fh [n] = fh [n-1])
Fh0 --;
}
If (sum> = 16 & ZM0> = 2 & zm0> = 2 & sz0> = 2 & fh0> = 2)
Printf ("% d Best \ n", sum );
Else if (sum> = 10 & ZM0> = 1 & zm0> = 1 & sz0> = 1 & fh0> = 1)
Printf ("% d Strong \ n", sum );
Else if (sum> = 8 & (ZM0> = 1 & zm0> = 1 & sz0> = 1) | (zm0> = 1 & ZM0> = 1 & fh0> = 1) | (zm0> = 1 & sz0> = 1 & fh0> = 1 )))
Printf ("% d Medium \ n", sum );
Else
Printf ("% d Weak \ n", sum );
Return 0;
}