Problem:
Score : 200 points

Problem Statement

You have a digit sequence S of length 4. You are wondering which of the following formats S is in:

YYMM format: the last two digits of the year and the two-digit representation of the month (example: 01 for January), concatenated in this order

MMYY format: the two-digit representation of the month and the last two digits of the year, concatenated in this order

If S is valid in only YYMM format, print YYMM; if S is valid in only MMYY format, print MMYY; if S is valid in both formats, print AMBIGUOUS; if S is valid in neither format, print NA.

Constraints

S is a digit sequence of length 4.

Input

Input is given from Standard Input in the following format:

S

Output

Print the specified string: YYMM, MMYY, AMBIGUOUS or NA.

Sample Input 1

1905

Sample Output 1

YYMM

May XX19 is a valid date, but 19 is not valid as a month. Thus, this string is only valid in YYMM format.

Sample Input 2

0112

Sample Output 2

AMBIGUOUS

Both December XX01 and January XX12 are valid dates. Thus, this string is valid in both formats.

Sample Input 3

1700

Sample Output 3

NA

Neither 0 nor 17 is valid as a month. Thus, this string is valid in neither format.

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <ctype.h>

void inp_str(int*);

int main(void) {

int input1[100];
inp_str(input1);

int S;
S=input1[0];

int a,b;
a=S/100;
b=S % 100;

if((a>12)&&(b<13)) {
	printf("YYMM\n");
} else if ((b>12)&&(a<13)) {
	printf("MMYY\n");
} else if (((b<13)||(a<13))&&(a>0)&&(b>0)) {
	printf("AMBIGUOUS\n");
} else {
	printf("NA\n");
}
    return 0;

}


void inp_str(int instr[]){
char *ptr;
int z,i,count;
char str[100]={'\0'};
char *ln;

fgets(str, 100, stdin); // 入力："123456\n"
ln = strchr(str, '\n'); /* 改行文字を検索 */
if (ln != NULL) { /* 改行が読み取られていたかどうか */
   *ln = '\0';    /* 改行文字を終端文字に置き換える */
}
else {            /* 入力ストリーム上に文字が残ってる場合 */
   while (1) {    /* 改行文字が読み取られるまで空読みする */
       z = getchar();
      if (z == '\n' || z == EOF) break;
   }
}
	count=0;
    ptr = strtok(str, " ");
    i = atoi(ptr);
    instr[count]=i;
    count=count+1;
    // 2回目以降
    while(ptr != NULL) {
        // strtok関数により変更されたNULLのポインタが先頭
        ptr = strtok(NULL, " ");
        // ptrがNULLの場合エラーが発生するので対処
        if(ptr != NULL) {
		    i = atoi(ptr);
    instr[count]=i;
    		count=count+1;
        }
    }
}