hdu 1019 Least Common Multiple

hdu 1019 Least Common Multiple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 29850 Accepted Submission(s): 11285

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

/*第一种：(推荐）

*/

#include<cstdio>

int gcd(int a,int b)

{

return !b?a:gcd(b,a%b);

}

int lcm(int a,int b)

{

return a*b/gcd(a,b); //a*b/gcd(a,b)就错了 ,可能是 a*b 的值过大 ，应定义为long long类型吧。

}

int main()

{

int T,i,n;

scanf("%d",&T);

while(T--)

{

int a[1003];

scanf("%d",&n);

if(n==1)

{

scanf("%d",&a[0]);

printf("%d\n",a[0]);

}

if(n>=2)

{

for(i=0; i<2; i++)

scanf("%d",&a[i]);

long long r=lcm(a[0],a[1]);

for(i=2; i<n; i++)

{

scanf("%d",&a[i]);

r=lcm(r,a[i]);

}

printf("%I64d\n",r);

}

}

return 0;

}

/*

第二种：

*/

#include<cstdio>

int gcd(int a,int b)

{

return !b?a:gcd(b,a%b);

}

int main()

{

int T,i,n;

scanf("%d",&T);

while(T--)

{

int a[1003];

scanf("%d",&n);

if(n==1)

{

scanf("%d",&a[0]);

printf("%d\n",a[0]);

}

if(n>=2)

{

for(i=0; i<n; i++)

{

scanf("%d",&a[i]);

}

for(i=1; i<n; i++)

{

a[i]=a[i-1]/gcd(a[i-1],a[i])*a[i]; //同上 a[i]=a[i-1]*a[i]/gcd(a[i-1],a[i]) oj会判错

}

printf("%d\n",a[n-1]);

}

}

return 0;

}