Theorem Content:
For any \ (a, b \ in Z \) and their greatest common divisor \ (D \) , on unknown \ (X \) and \ (Y \) is the linear equation variable \ (ax + by = c \ ) Solutions of integers \ ((x, y) \ ) if and only if \ (D | C \) , there are infinitely many solutions found. In particular, there must be an integer that the \ (ax + by = d \ ) established
inference:
\ (a, b \) coprime integers iff \ (x, y \) makes \ (= AX +. 1 by \) .
prove:
Set \ (gcd (a, b)
= d \) readily available: \ (D | A \) , \ (D | B \)
and \ (\ Because \) \ (X \ in the Z, Y \ in the Z \ )
then available: \ (d | AX + by \)
Like syndrome finished
About this question is to be extended to more than two variables variables
beg \ (n-1 \) times \ (gcd \) to
the code is simple home
#include <cstdio>
#include <iostream>
using namespace std;
int n, ans, x, y;
int read() {
int s = 0, w = 1;
char ch = getchar();
while(!isdigit(ch)) {if(ch == '-') w = -1; ch = getchar();}
while(isdigit(ch)) {s = s * 10 + ch - '0'; ch = getchar();}
return s * w;
}
int gcd(int x, int y) {
// if(x < 0) x = -x;
// if(y < 0) y = -y;
return y == 0 ? x : gcd(y, x % y);
}
int main() {
n = read();
x = read(), y = read();
if(x < 0) x = -x;
if(y < 0) y = -y;
ans = gcd(x, y);
for(int i = 1; i <= n - 2; i++) {
y = read();
if(y < 0) y = -y;
ans = gcd(ans, y);
// x = ans;
}
cout << ans << endl;
return 0;
}Thank you for watching, I wish good health!