扩展欧几里得例题(luogu_1082)

luo gu

由$a\ast x\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}b)$ 推导为扩展欧几里得

• ->$a\ast x\mathrm{m}\mathrm{o}\mathrm{d}b==1\mathrm{m}\mathrm{o}\mathrm{d}b$
• ->$a\ast x\mathrm{m}\mathrm{o}\mathrm{d}b=1$
• 即-> $a\ast x=n\ast b+1$(n是常数)
• -> $a\ast x-n\ast b=1$
• ->$a\ast x+y\ast b=1$ (y = -n,常数无影响)
  模板：

#include <iostream>
#include <string.h>
#include <queue>

using namespace std;

int exgcd(int a,int b,long long & x,long long & y) {
	if (b == 0) {
		x = 1;
		y = 0;
		return a;
	}
	int res = exgcd(b,a%b,x,y);
	// 回溯的时候进行  推倒x,y
	long long temp = y;
	y = x - (a/b)*y;
	x = temp;
	return res;
}


int main() {
	int a,b;
	long long x,y;
	cin >> a >> b;
	exgcd(a,b,x,y);
	if (x > 0) {
		while (x > 0) 
			x -= abs(b);
		x += abs(b);
		cout << x << endl;
	}
	else {
		while (x < 0) 
			x += abs(b);
		cout << x << endl;
	}
	return 0;
}