求最小公倍数 编程拜托各位大神Least Common Multiple Time Limit: 1 Second Memory Limit: 32768 KB The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the s
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求最小公倍数 编程拜托各位大神Least Common Multiple Time Limit: 1 Second Memory Limit: 32768 KB The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the s
求最小公倍数 编程拜托各位大神
Least Common Multiple Time Limit: 1 Second Memory Limit: 32768 KB The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the set. For example, the LCM of 5, 7 and 15 is 105. Input Input will consist of multiple problem instances. The first line of the input will contain a single integer indicating the number of problem instances. Each instance will consist of a single line of the form m n1 n2 n3 ... nm where m is the number of integers in the set and n1 ... nm are the integers. All integers will be positive and lie within the range of a 32-bit integer. Output For each problem instance, output a single line containing the corresponding LCM. All results will lie in the range of a 32-bit integer. Sample Input 2 3 5 7 15 6 4 10296 936 1287 792 1 Sample Output 105 10296 求几组数的的最小公倍数,前面不知怎么输入,请指点,尽量用pascal语言,不行就用c语言,要说明,谢谢
求最小公倍数 编程拜托各位大神Least Common Multiple Time Limit: 1 Second Memory Limit: 32768 KB The least common multiple (LCM) of a set of positive integers is the smallest positive integer which is divisible by all the numbers in the s
Var n,i,j,m,a,z:longint; function gcd(a,b:longint):longint;//求最大公约数函数 Begin if a mod b = 0 Then exit(b) else exit(gcd(b,a mod b)); End; Begin readln(n); For i:=1 to n do Begin z:=1; while not(eoln) do Begin//若输入未换行则继续读入 read(a); z:=z*a div gcd(z,a);//由lcm(a,b)=a*b/gcd(a,b)得到（最小公倍数等于两数之和除以两数最大公约数） End; readln;//将换行符读入 writeln(z); End; End.