least common multiple

time limit: 2000/1000 ms (java/others) memory limit: 65536/32768 k (java/others)total submission(s): 64855 accepted submission(s): 24737

problem description

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 

这道水题连续提交五次，第六次才ac。。。。。。（我真是菜，最近做做的题有点自闭）

这是是我多次修改仍然不ac的题解（测试样例全过自己测了几个也没问题，很绝望）

#include <stdio.h> #include <math.h>  int main() {     int t,temp;     scanf("%d",&t);     for(int i=1; i <= t; i++) {         int n;         scanf("%d",&n);         for(int j=1; j <= n ; j++) {             int num;             scanf("%d",&num);             if(j == 1) {                 temp=num;                 continue;             }             int max;             if(temp > num)                 max=num;             else                 max=temp;             for( ; max >= 1; max--)                 if(temp%max == 0&&num%max == 0)                     break;             temp=temp*num/max;         }         printf("%dn",temp);     }     return 0; } 

最后有dalao指点，把求公因数的方法改成辗转相乘法并用函数嵌套终于ac 无奈绝望╮(╯﹏╰）╭ 关于辗转相除法求最大公因数，举个栗子： a=6, b=4; 第一步 a=a%b=6%4=2； ​ 第二步 b=4； ​ 第三步 a%b=2%4=0，此时a == 0，b=(上一步)a; ​ 此时b就是最大公因数； （算了上详细讲解链接 ）

以下为正确代码

#include <stdio.h> #include <math.h>  int gcd(int a,int b){   //辗转相除法          if(b == 0)     return a;     return gcd(b,a%b); }  int lcm(int a,int b) {     return a/gcd(a,b)*b;    //把函数打包 }  int main() {     int t,temp,n;     scanf("%d",&t);     for(int i=1; i <= t; i++) {         scanf("%d",&n);         int temp = 1;         for(int j=1; j <= n ; j++) {             int num;             scanf("%d",&num);             temp = lcm(temp,num);         }         printf("%dn",temp);     }     return 0; }