题目链接：http://www.spoj.com/problems/HG/
题目大意：分别给N个和M个小于等于1e9的数，前N个数相乘为A，后M个数相乘为B，问你GCD(A, B) 1<= N<= 1000，1 <=M <= 1000
解题思路：枚举a[i] 和b[i]，计算两者GCD，然后结果乘以GCD，两者都除以GCD。最后判断一下如何输出即可。
代码：
1 const int maxn = 1e4 + 5;
2 int n, m;
3 ll a[maxn], b[maxn];
4
5 int gcd(int x, int y){
6 return y == 0? x: gcd(y, x % y);
7 }
8 void solve(){
9 ll ans = 1;
10 bool flag = false;
11 for(int i = 0; i < n; i++){
12 for(int j = 0; j < m; j++){
13 int tmp = gcd(a[i], b[j]);
14 a[i] /= tmp; b[j] /= tmp;
15 ans *= tmp;
16 if(ans >= mod){
17 flag = true;
18 ans %= mod;
19 }
20 }
21 }
22 if(flag) printf("%09lld\n", ans % mod);
23 else printf("%lld\n", ans);
24 }
25
26 int main(){
27 scanf("%d", &n);
28 for(int i = 0; i < n; i++) scanf("%lld", &a[i]);
29 scanf("%d", &m);
30 for(int i = 0; i < m; i++) scanf("%lld", &b[i]);
31 solve();
32 }
题目：
HG - HUGE GCD
RK has received a homework assignment to compute the greatest common divisor of the two positive integers Aand B. Since the numbers are quite large, the professor provided him with N smaller integers whose product is A, and M integers with product B.
RK would like to verify his result, so he has asked you to write a program to solve his problem. If the result is more than 9 digits long, output only the last 9 digits.
INPUT
The first line of input contains the positive integer N (1<= N<= 1000).
The second line of input contains N space-separated positive integers less than 10^9, whose product is the number A.
The third line of input contains the positive integer M (1 <=M <= 1000).
The fourth line of input contains M space-separated positive integers less than 10^9, whose product is the number B.
OUTPUT
The first and only line of output must contain the greatest common divisor of numbers A and B. If the result is more than 9 digits long, output only the last (least significant) 9 digits.
SAMPLE
Input
3
2 3 5
2
4 5
Output
10
Input
3
358572 83391967 82
3
50229961 1091444 8863
Output
000012028
First sample description: The greatest common divisor of numbers A = 30 and B = 20 equals 10.