PAT B1062

来源:互联网 发布:芮纳纪 淘宝 编辑:程序博客网 时间:2024/06/07 17:47
#include <stdio.h>int gcd(int a,int b){return b==0?a:gcd(b,a%b);}int main(){  int N,M,N1,M1,K,val,val1,count=0,i,A[1000];  double s,s1,max,min;  scanf("%d/%d %d/%d %d",&N,&M,&N1,&M1,&K);  val=gcd(N,M);  s=(N*1.0/val)/(M*1.0/val);  val1=gcd(N1,M1);  s1=(N1*1.0/val1)/(M1*1.0/val1);  max=s>s1?s:s1;  min=s<s1?s:s1;  for(i=0;i<1000;i++)    if(i*1.0/K>min&&i*1.0/K<max)    {  if(1==gcd(i,K))          A[count++]=i;    }  for(i=0;i<count;i++)  {  printf("%d/%d",A[i],K);    if(i<count-1)      printf(" ");  }  printf("\n");return 0;}
0 0
原创粉丝点击