LightOJ 1024 - Eid (高精度乘法求n个数的LCM)
In a strange planet there are n races. They are completely different as well as their food habits. Each race has a food-eating period. That means the ith race eats after every xi de-sec (de-sec is the unit they use for counting time and it is used for both singular and plural). And at that particular de-sec they pass the whole day eating.
The planet declared the de-sec as 'Eid' in which all the races eat together.
Now given the eating period for every race you have to find the number of de-sec between two consecutive Eids.
Input
Input starts with an integer T (≤ 225), denoting the number of test cases.
Each case of input will contain an integer n (2 ≤ n ≤ 1000) in a single line. The next line will contain n integers separated by spaces. The ithinteger of this line will denote the eating period for the ith race. These integers will be between 1 and 10000.
Output
For each case of input you should print a line containing the case number and the number of de-sec between two consecutive Eids. Check the sample input and output for more details. The result can be big. So, use big integer calculations.
Sample Input
2
3
2 20 10
4
5 6 30 60
Sample Output
Case 1: 20
Case 2: 60
题意：给出n个数，求出这n个数的最小公倍数
思路：把每个数都拆分成素因子的乘积，记录每个素因子的最大数量值，然后相乘
例如 ： 5 = 5 一个 5
25 = 5 * 5 两个 5
30=2 * 3 * 5 一个 2 ， 一个 3 ，一个 5
60=2 * 2 * 3 * 5 俩个 2 ，一个 3 ，一个 5
可以看出 2的最大数量值为2 ，3的最大数量数量值为1，5的最大数量值为2，所以lcm=2*2*3*5*5；
由于 n 的范围为1000，值小于10000，所以求出来的值会非常大，用高精度乘法来做
代码：具体看代码
#include
#include
#include
using namespace std;
typedef unsigned long long ll;
const int INF=0x3f3f3f3f;
int ans[1010]; //存lcm的值 数组中每位存的是四位数
int prime[10010];
int m[10010];
void isprime() //素数筛
{
for(int i=0;i<=10000;i++)
prime[i]=1;
prime[0]=prime[1]=0;
for(int i=2;i<=10000;i++)
if(prime[i])
for(int j=i+i;j<=i;j+=i)
prime[j]=0;
}
void mul(int ans[],int v) //高精度乘法------万进制
{
for(int i=0;i<=1000;i++) //模拟乘法
ans[i]=ans[i]*v;
for(int i=0;i<=1000;i++)
{
ans[i+1]+=ans[i]/10000;
ans[i]%=10000;
}
}
void print() //输出
{
int pos=1000;
while(pos>=0&&ans[pos]==0) //去除前导零
pos--;
printf("%d",ans[pos--]);
while(pos>=0)
printf("%04d",ans[pos--]);
printf("\n");
}
int main()
{
int t,n,x,K=1;
isprime();
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
memset(m,0,sizeof(m));
memset(ans,0,sizeof(ans));
for(int i=0;i