[백준 16017] Telemarketer or not? :: C언어

문제

Here at the Concerned Citizens of Commerce (CCC), we have noted that telemarketers like to use seven-digit phone numbers where the last four digits have three properties. Looking just at the last four digits, these properties are:

• the first of these four digits is an 8 or 9;
• the last digit is an 8 or 9;
• the second and third digits are the same.

For example, if the last four digits of the telephone number are 8229, 8338, or 9008, these are telemarketer numbers.

Write a program to decide if a telephone number is a telemarketer number or not, based on the last four digits. If the number is not a telemarketer number, we should answer the phone, and otherwise, we should ignore it.

입력

The input will be 4 lines where each line contains exactly one digit in the range from 0 to 9.

출력

Output either ignore if the number matches the pattern for a telemarketer number; otherwise, output answer.

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처음에 if문의 중첩으로 문제를 해결했는데, 실행은 제대로 되지만 틀렸다는 결과가 나왔다.

그래서 논리연산자 &&를 사용해서 문제를 해결했다.

 

if문 중첩도 맞는데...

 

---

#include<stdio.h>

int main() {
	int arr[4];
	for (int a = 0; a < 4; a++) {
		scanf("%d", &arr[a]);
	}
	if (arr[0] > 7 && arr[1] == arr[2] && arr[3] > 7) printf("ignore\n");
	else printf("answer\n");
	return 0;
}

---

틀렸지만 if문 중첩으로 푼 알고리즘 (if문 중첩에도 논리연산자 ||가 사용되었다)

#include<stdio.h>

int main() {
	int arr[4];
	for (int a = 0; a < 4; a++) {
		scanf("%d", &arr[a]);
	}
	if (arr[0] == 8 || arr[0] == 9) {
		if (arr[3] == 8 || arr[3] == 9) {
			if (arr[1] == arr[2]) {
				printf("ignore\n");
			}
		}
	}
	else printf("answer\n");
	return 0;
}