Stupid bear (Nanyang oj62) and bear Nanyang oj62

Stupid bear (Nanyang oj62) and bear Nanyang oj62
Stupid bear time limit: 2000 MS | memory limit: 65535 KB difficulty: 2

Description

Dummies have a small vocabulary, so they have a headache every time they make English choices. However, he found a method. The experiment proves that it is very likely to select the right option when using this method!
The specific description of this method is as follows: Suppose maxn is the number of occurrences of the most frequently-occurring letters in a word, and minn is the number of occurrences of the letters with the least number of occurrences in a word, if maxn-minn is a prime number, the stupid bear thinks it is a Lucky Word. Such words are probably the correct answer.

Input

The first row of data N (0 <N <100) indicates the number of test data groups.
Each group of test data has only one line of input, which is a word. Only lowercase letters may appear and the length is less than 100.

Output

Two rows are output for each group of test data. The first row is a string. If the input Word is Lucky Word, "Lucky Word" is output; otherwise, "No Answer" is output ";
The second line is an integer. If the input Word is Lucky Word, the value of maxn-minn is output; otherwise, 0 is output.

Sample Input

2errorolympic

Sample output

Lucky Word2No Answer0

# Include <stdio. h> # include <string. h ># include <algorithm> using namespace std; int s [200] = {, 0}; int main () {// prime number table. Int I, j; for (I = 2; I * I <200; I ++) {if (! S [I]) {for (j = I + I; j <200; j + = I) {s [j] = 1 ;}} char a [200]; int B [200]; int len, k, max, min, test; scanf ("% d", & test); getchar (); while (test --) {memset (B, 0, sizeof (B); gets (a); len = strlen (a); for (I = 0; I <len; I ++) // count the number of occurrences of each character. {For (j = 0; j <len; j ++) {if (a [I] = a [j]) B [I] ++ ;}} max = min = B [0]; for (I = 0; I <len; I ++) // find the characters that appear most frequently and least. {If (B [I]> max) max = B [I]; if (B [I] <min) min = B [I];} k = max-min; if (! S [k]) {printf ("Lucky Word \ n"); printf ("% d \ n", k );} else {printf ("No Answer \ n"); printf ("0 \ n") ;}} return 0 ;}

I think there is no problem with the program, but I cannot submit it. # include <stdioh>

What about your question description? The main function is a bit dizzy and you don't know what to do. Give the questions and test examples. It's meaningless to guess your logic.

If (! (Check % 2 ))
If (check! = 2) return 0;
Else return 1;

Whether
If (! (Check % 2 ))
{
If (check! = 2) return 0;
Else return 1;
}

Nanyang Institute of Technology acm stupid bear problem, I feel there is no problem with the program, but an error occurs when I submit

What programs are you using? Why are there so many cases?
Why not directly use B [a [I]-'a'] ++