Difference between revisions of "Solução: Problema da simplificação das frações (Bruno Barbosa)"
From AdonaiMedrado.Pro.Br
Line 24: | Line 24: | ||
while(n || m){ | while(n || m){ | ||
− | if(!n || !m) | + | if(!n || !m){ |
− | printf("%d %d\n", n, m); | + | |
+ | if(n != 0) | ||
+ | printf("%d %d\n", n/n, m); | ||
+ | if(m != 0) | ||
+ | printf("%d %d\n", n, m/m); | ||
+ | } | ||
else | else |
Latest revision as of 18:53, 12 June 2009
#include <stdio.h> int analisaImprime(int num, int x, int y); int analisaImprime(int i, int x, int y){ while((x%i!=0)||(y%i!=0)){ i--; } if((x%i == 0)&&(y%i==0)) printf("%d %d\n", x/i, y/i); else printf("%d %d\n", x, y); } int main(){ int n, m; scanf("%d %d", &n, &m); while(n || m){ if(!n || !m){ if(n != 0) printf("%d %d\n", n/n, m); if(m != 0) printf("%d %d\n", n, m/m); } else analisaImprime(n, n, m); scanf("%d %d", &n, &m); } }