1 条题解
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0
C :
#include <stdio.h> int gcd(int m, int n){ return (n>0) ? gcd(n, m%n) : m; } int lcm(int m, int n){ return m * n / gcd(m,n); } int main(){ int n,a,b,c; scanf("%d",&n); while(n--){ scanf("%d %d",&a, &b); printf("%d %d\n", gcd(a, b), lcm(a, b)); } return 0; }
C++ :
#include<iostream> using namespace std; int gcd(int a,int b){ if(b==0) return a; else return gcd(b,a%b); } int main(){ int n; cin>>n; while(n--){ int a,b,max,min; cin>>a>>b; min=gcd(a,b); max=a*b/min; cout<<min<<" "<<max<<endl; } }
Pascal :
program x1; var t,a,b,i,m,n,r:integer; d:real; begin readln(t); for i:=1 to t do begin readln(a,b); n:=a; m:=b; r:=n mod m; while r<>0 do begin n:=m; m:=r; r:=n mod m; end; d:=a*b/m; writeln(m,' ',d:0:0); end; end.
Java :
import java.util.Scanner; public class Main { public static void main(String[] args) { Scanner scan=new Scanner(System.in); int n=scan.nextInt(); while(n-->0) { int i=scan.nextInt(); int j=scan.nextInt(); int x=i,y=j; if(i>=j){ int a=i%j; while(a!=0){ i=j; j=a; a=i%j; } System.out.print(j+" "); }else{ int t; t=j; j=i; i=t; int a=i%j; while(a!=0){ i=j; j=a; a=i%j; } System.out.print(j+" "); } System.out.println((x*y)/j); } } }
- 1
信息
- ID
- 2006
- 时间
- 1000ms
- 内存
- 128MiB
- 难度
- (无)
- 标签
- 递交数
- 0
- 已通过
- 0
- 上传者