[백준/C언어/C99] 5086번 배수와 약수 | 단계별로 풀어보기

1. 문제

https://www.acmicpc.net/problem/5086

 

2. 풀이

#include <stdio.h>

int main(void){
    int tests[10001][2];
    int results[10001];
    int count = 0;
    int flag = 0;

    char factor[] = "factor\0";
    char multiple[] = "multiple\0";
    char neither[] = "neither\0";

    while(1){
        scanf("%d %d", &tests[count][0], &tests[count][1]);
        if(!tests[count][0] && !tests[count][1]){
            results[count] = -1;
            break; 
        }

        if(tests[count][1] % tests[count][0] == 0){
            results[count] = 0;
        } else if (tests[count][0] % tests[count][1] == 0){
            results[count] = 1;
        } else {
            results[count] = 2;
        }

        count++;
    }

    for(int i = 0; i < count+1; i++){
        switch (results[i])
        {
        case 0:
            printf("%s\n", factor);
            break;
        case 1:
            printf("%s\n", multiple);
            break;
        case 2:
            printf("%s\n", neither);
            break;
        default:
            printf("");
            break;
        }
    }
}