hdu1019 gcd和lcm

Least Common Multiple

http://acm.hdu.edu.cn/showproblem.php?pid=1019

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

解题思路
经典的最大公约数（gcd）和最小公倍数（lcm）题目。
这两个算法是代码模块。

注意：
先除后乘防止数据溢出！！

#include <stdio.h>
#include <stdlib.h>

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

int lcm(int a, int b)
{
    int g;

    g = gcd(a, b);
    return a / g * b;
}

int main (int argc, char const* argv[])
{
    int c, n, i, ans, t;

    scanf("%d", &c);

    while (c--) {
        scanf("%d", &n);

        scanf("%d", &ans);
        for (i = 1; i < n; i++) {
            scanf("%d", &t);
            ans = lcm(ans, t);
        }

        printf("%d\n", ans);
    }

    return 0;
}