Difference between revisions of "Solução: Problema da simplificação das frações (Bruno Barbosa)"
From AdonaiMedrado.Pro.Br
Line 4: | Line 4: | ||
int main(){ | int main(){ | ||
− | int n, m, i; | + | int n, m, i, teste; |
scanf("%d %d", &n, &m); | scanf("%d %d", &n, &m); | ||
Line 20: | Line 20: | ||
if(n>m){ | if(n>m){ | ||
i = m; | i = m; | ||
− | printf("%d %d\n", n/i, m/i); | + | while((m%i!=0)&&(n%i!=0)){ |
+ | i--; | ||
+ | } | ||
+ | if((m%i == 0)&&(n%i==0)) | ||
+ | printf("%d %d\n", n/i, m/i); | ||
+ | |||
+ | else | ||
+ | printf("%d %d\n", n, m); | ||
+ | |||
} | } | ||
− | + | ||
− | if( | + | if(n<m){ |
− | i= n; | + | i = n; |
− | printf("%d %d\n", n/i, m/i); | + | while((m%i!=0)&&(n%i!=0)){ |
+ | i--; | ||
+ | } | ||
+ | if((m%i!=0)&&(n%i!=0)) | ||
+ | printf("%d %d\n", n/i, m/i); | ||
+ | else | ||
+ | printf("%d %d\n", n, m); | ||
} | } | ||
+ | |||
} | } | ||
scanf("%d %d", &n, &m); | scanf("%d %d", &n, &m); | ||
− | } | + | } |
} | } | ||
− | |||
− | |||
− | |||
</code> | </code> |
Revision as of 19:10, 5 June 2009
#include <stdio.h> int main(){ int n, m, i, teste; scanf("%d %d", &n, &m); while(n || m){ if(!n || !m){ printf("%d %d\n", n, m); } else{ if(n==m){ i = m; printf("%d %d\n", n/i, m/i); } if(n>m){ i = m; while((m%i!=0)&&(n%i!=0)){ i--; } if((m%i == 0)&&(n%i==0)) printf("%d %d\n", n/i, m/i); else printf("%d %d\n", n, m); } if(n<m){ i = n; while((m%i!=0)&&(n%i!=0)){ i--; } if((m%i!=0)&&(n%i!=0)) printf("%d %d\n", n/i, m/i); else printf("%d %d\n", n, m); } } scanf("%d %d", &n, &m); } }