codeforces 664A A. Complicated GCD(水题)

题目链接:

A. Complicated GCD

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

 

Greatest common divisor GCD(a, b) of two positive integers a and b is equal to the biggest integer d such that both integers a and bare divisible by d. There are many efficient algorithms to find greatest common divisor GCD(a, b), for example, Euclid algorithm.

Formally, find the biggest integer d, such that all integers a, a + 1, a + 2, ..., b are divisible by d. To make the problem even more complicated we allow a and b to be up to googol, 10100 — such number do not fit even in 64-bit integer type!

Input

 

The only line of the input contains two integers a and b (1 ≤ a ≤ b ≤ 10100).

Output

 

Output one integer — greatest common divisor of all integers from a to b inclusive.

Examples

 

input

1 2

output

1

input

61803398874989484820458683436563811772030917980576 61803398874989484820458683436563811772030917980576

output

61803398874989484820458683436563811772030917980576

题意：

[a,b]的最大公约数是多少;

思路：

可以发现,a和b不相同时的最大公约数是1,相同时是其本身;

AC代码：

/*2014300227    664A - 9    GNU C++11    Accepted    15 ms    4 KB*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e4+5;
typedef long long ll;
const int mod=1e9+7;
char str1[110],str2[110];
int check()
{
    int len1=strlen(str1),len2=strlen(str2);
    if(len1==len2)
    {
        for(int i=0;i<len1;i++)
        {
            if(str1[i]!=str2[i])return 0;
        }
        return 1;
    }
    return 0;
}
int main()
{
    scanf("%s%s",str1,str2);
    if(check())printf("%s
",str1);
    else printf("1
");



    return 0;
}