Difference between revisions of "Solução: Problema da simplificação das frações (Bruno Barbosa)"
From AdonaiMedrado.Pro.Br
Line 20: | Line 20: | ||
if(n>m){ | if(n>m){ | ||
i = m; | i = m; | ||
− | while((m%i!=0) | + | while((m%i!=0)||(n%i!=0)){ |
i--; | i--; | ||
} | } | ||
Line 33: | Line 33: | ||
if(n<m){ | if(n<m){ | ||
i = n; | i = n; | ||
− | while((m%i!=0) | + | while((m%i!=0)||(n%i!=0)){ |
i--; | i--; | ||
} | } | ||
− | if((m%i | + | if((m%i==0)&&(n%i==0)) |
printf("%d %d\n", n/i, m/i); | printf("%d %d\n", n/i, m/i); | ||
else | else |
Revision as of 21:41, 5 June 2009
#include <stdio.h> int main(){ int n, m, i; 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); } }