Problem:
Score : 300 points

Problem Statement

There are N monsters, numbered 1, 2, ..., N.

Initially, the health of Monster i is A_i.

Below, a monster with at least 1 health is called alive.

Until there is only one alive monster, the following is repeated:

A random alive monster attacks another random alive monster.

As a result, the health of the monster attacked is reduced by the amount equal to the current health of the monster attacking.

Find the minimum possible final health of the last monster alive.

Constraints

All values in input are integers.

2 \leq N \leq 10^5

1 \leq A_i \leq 10^9

Input

Input is given from Standard Input in the following format:

N
A_1 A_2 ... A_N

Output

Print the minimum possible final health of the last monster alive.

Sample Input 1

4
2 10 8 40

Sample Output 1

2

When only the first monster keeps on attacking, the final health of the last monster will be 2, which is minimum.

Sample Input 2

4
5 13 8 1000000000

Sample Output 2

1

Sample Input 3

3
1000000000 1000000000 1000000000

Sample Output 3

1000000000

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

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

#define i_cin(X) scanf("%d", &X)
#define i_cin2(X,Y) scanf("%d %d",&X,&Y)
#define i_cin3(X,Y,Z) scanf("%d %d %d",&X,&Y,&Z)
#define l_cin(X) scanf("%ld",&X)
#define l_cin2(X,Y) scanf("%ld %ld",&X,&Y)
#define s_cin(X) scanf("%s",&X)

#define  i_cout(X) printf("%d\n",X)
#define  f_cout(X) printf("%g\n",X)
#define  i_cout2(X,Y) printf("%d %d\n",X,Y)
#define  l_cout(X) printf("%ld\n",X)
#define  s_cout(X) printf("%s\n",X)
#define  ll_cout(X) printf("%lld\n",X)

typedef long long int lli;
#define rept(x, s, n) for (int x = s; x < n; x++)

#define YES_RTN s_cout("Yes");return 0
#define NO_RTN  s_cout("No");return 0


int i_cins(int n, int* A);
int l_cins2(int n, long* A,long* B);
int s_dsort(const void* a, const void* b);
int s_asort(const void* a, const void* b);


int main(void) {

	int    A[100001], N, min, tmp, i;
	i_cin(N);
	i_cins(N, A);

	qsort(A, N, sizeof(int), s_asort);

	min = A[0];
	for (i = N - 1; i > 0; i--) {
		tmp = A[i] % min;
		if (min > tmp && tmp) min = tmp;
		A[i] = tmp;
	}

	qsort(A, N, sizeof(int), s_asort);

//	for (i = N - 1; i >= 0; i--) {
//		if (A[i] == 0)break;
//		i_cout(A[i]);
//		s_cout("min:\n", min);
//	}
	if (!A[N - 2]) { i_cout(min); return 0; }

//	for (i = N - 1; i >= 0; i--) {
//		if (A[i] == 0)break;
//		i_cout(A[i]);
//	}

	while (1) {

		for (i = N - 1; i >= 0; i--) {
			if (A[i] == 0)break;
			tmp = A[i] % min;
			if (min > tmp && tmp) min = tmp;
			A[i] = tmp;
		}
		qsort(A, N, sizeof(int), s_asort);
	
//		for (i = N - 1; i >= 0; i--) {
//			if (A[i] == 0)break;
//			i_cout(A[i]);
//			s_cout("min:\n",min);
//		}
		if (!A[N - 2]) break;

	}
	i_cout(min);
	return 0;
	}
	
//降順
int s_dsort(const void* a, const void* b) {
	return(*(int*)b - *(int*)a);
}
int s_asort(const void* a, const void* b) {
	return(*(int*)a -*(int*)b);
}
//
int l_cins2(int n, long* a, long* b) {
	int i;
	rept(i, 0, n) {
		l_cin2(*(a + i), *(b + i));
	}
	return 0;
}
int i_cins(int n, int* a) {
	int i;
	rept(i, 0, n) {
		i_cin(*(a + i));
	}
	return 0;
}