杭电acm：最小公倍数（附源码）

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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

23 5 7 156 4 10296 936 1287 792 1

Sample Output

10510296

思路：写一个两两求最小公倍数的函数（用long long）然后循环处理

ps vc环境的__int64貌似在处理较小的数时容易出现错误

#include<stdio.h>#include<iostream>using namespace std;long long  lcd(long long  n,long long  m){    long long r;    //scanf("%d%d",&n,&m);    if(n<m)swap(n,m);    long long p=n*m;    while(m!=0)    {        r=n%m;        n=m;        m=r;    }    return p/n;}int main(){    long long num[100];    int n1,n2,o;    scanf("%d",&n1);    while(n1--)    {        scanf("%d",&n2);        for(int o=0;o<n2;o++)        {            scanf("%lld",&num[o]);        }        for(o=1;o<n2;o++)        {            num[o]=lcd(num[o],num[o-1]);        }        printf("%lld\n",num[o-1]);    }}

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