扩展欧几里得 exGCD

Elementary Number Theory - Extended Euclid Algorithm

Time Limit : 1 sec, Memory Limit : 65536 KB 
Japanese version is here

Extended Euclid Algorithm

Given positive integers a and b, find the integer solution (x, y) to ax+by=gcd(a,b), where gcd(a,b) is the greatest common divisor of a and b.

Input

a b

Two positive integers a and b are given separated by a space in a line.

Output

Print two integers x and y separated by a space. If there are several pairs of such x and y, print that pair for which |x|+|y| is the minimal (primarily) and x ≤ y (secondarily).

Constraints

• 1 ≤ a, b ≤ 109

Sample Input 1

4 12

Sample Output 1

1 0

Sample Input 2

3 8

Sample Output 2

3 -1

#include <bits/stdc++.h>

#define fread_siz 1024

inline int get_c(void)
{
    static char buf[fread_siz];
    static char *head = buf + fread_siz;
    static char *tail = buf + fread_siz;

    if (head == tail)
        fread(head = buf, 1, fread_siz, stdin);

    return *head++;
}

inline int get_i(void)
{
    register int ret = 0;
    register int neg = false;
    register int bit = get_c();

    for (; bit < 48; bit = get_c())
        if (bit == '-')neg ^= true;

    for (; bit > 47; bit = get_c())
        ret = ret * 10 + bit - 48;

    return neg ? -ret : ret;
}

int exgcd(int a, int b, int &x, int &y)
{
    if (!b)
    {
        x = 1;
        y = 0;
        return a;
    }
    int ret = exgcd(b, a%b, y, x);
    y = y - x * (a / b);
    return ret;
}

signed main(void)
{
    int x, y;
    int a = get_i();
    int b = get_i();
    exgcd(a, b, x, y);
    printf("%d %d
", x, y);
}

@Author: YouSiki