hdoj1019 Least Common Multiple

题目：

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

题目意思很简单，就是给你一组序列，让你求出这组序列的最小公倍数。

这道题有两个注意的地方：

1. 考虑这个序列中只有一个元素的情况（最小公倍数就是那个数）。
2. 当心溢出（用long long）。

参考代码：

#include 
#include 
#include 
#include 
using namespace std;
typedef long long ll;

vector v;

void init() {
    v.clear();
}

void input(int n) {
    ll num;
    for (int i = 0;i < n;++i) {
        cin >> num;
        v.push_back(num);
    }
}

ll gcd(ll a, ll b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}

ll lcm(ll a, ll b) {
    ll g = gcd(a, b);
    ll d = a * b;
    return d / g;
}

ll result(int n) {
    if (n == 1) return v[0];
    ll ans = lcm(v[0], v[1]);
    for (string::size_type i = 2;i < v.size();++i) {
        ans = lcm(ans, v[i]);
    }
    return ans;
}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        init();
        input(n);
        ll ans = result(n);
        cout << ans << endl;
    }
    return 0;
}