中国铁道建设出版
}
4-4 for(i=0;i<49;i++) { if(a[i]!=0) { } cout<<endl; cout<<setw(6)<<a[i]; n++; if(n%5==0) cout<<endl; }
#include <iostream.h>
int gcd(int,int);
int lcm(int,int);
void main()
{
int num1,num2,den1,den2,a,l,g; cin>>num1>>den1>>num2>>den2; cout<<num1<<"/"<<den1<<" , "<<num2<<"/"<<den2<<'\n'; cout<<"求分母的最小公倍数: "; l=lcm(den1,den2); cout<<l<<'\n';
cout<<"两分数的和为：";
a=(l/den1)*num1+(l/den2)*num2;
cout<<a<<'/'<<l<<'\n';
}
int gcd(int a,int b)
{
int i; for(i=a;i>=1;i--) if(!((a%i)||(b%i))) break; return (i); cout<<"求分子分母的最大公约数："; g=gcd(a,l); cout<<g<<'\n'; cout<<"输出最简分数："; cout<<a/g<<"/"<<l/g<<'\n';