Codeforces Round #347 (Div. 2)-Complicated GCD(字符串判等)

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

思路：
  这道题目第一眼看就感觉很难，但是认真看完题目后就能找到一个突破口，那就是[L,R]的所有最大公约数，因为相邻两个肯定互质，所以只要判断是否相等就可以了，相等就是本身，否则是1.

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<cstdio>
#include<vector>
using namespace std;
typedef long long ll;
#define T 100000 + 50

int main()
{
#ifdef zsc
freopen("input.txt","r",stdin);
#endif

int n,m,i,j,k;
string s,ss;
while(cin >> s >> ss)
{
if(s==ss)cout << s << endl;
else cout << 1 << endl;
}

return 0;
}