Difference between revisions of "Solução: Problema da simplificação das frações (Bruno Barbosa)"
From AdonaiMedrado.Pro.Br
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#include <stdio.h> | #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 main(){ | ||
| − | int n, m | + | |
| + | int n, m; | ||
scanf("%d %d", &n, &m); | scanf("%d %d", &n, &m); | ||
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while(n || m){ | while(n || m){ | ||
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if(!n || !m){ | if(!n || !m){ | ||
| − | + | ||
| − | + | if(n != 0) | |
| − | + | printf("%d %d\n", n/n, m); | |
| − | if(n | + | if(m != 0) |
| − | + | printf("%d %d\n", n, m/m); | |
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| − | if( | + | |
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} | } | ||
| + | |||
| + | else | ||
| + | analisaImprime(n, n, m); | ||
| + | |||
scanf("%d %d", &n, &m); | scanf("%d %d", &n, &m); | ||
} | } | ||
} | } | ||
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</code> | </code> | ||
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); } }
