Acwing-203-同余方程(扩展欧几里得)

链接:

https://www.acwing.com/problem/content/205/

题意:

求关于x的同余方程 ax ≡ 1(mod b) 的最小正整数解。

思路:

首先:扩展欧几里得推导.
有ax+by = gcd(a, b) = gcd(b, a%b),
ax+by = bx+(a%b)y
ax+by = bx+(a-(a/b)b)y
ax+by = bx + ay-(a/b)by
ax+by = ay + b(x-a/by)
有x' = y, y' = x-a/by
递归求解
对于ax = 1 (mod b).有b | ax+1. 令 ax+1 = -yb.
有ax+by = 1.用扩展欧几里得可以求出一个解.

代码:

#include <bits/stdc++.h>
using namespace std;

int ExGcd(int a, int b, int &x, int &y)
{
    if (b == 0)
    {
        x = 1, y = 0;
        return a;
    }
    int d = ExGcd(b, a%b, x, y);
    int tmp = y;
    y = x-(a/b)*y;
    x = tmp;
    return d;
}

int main()
{
    int a, b, x, y;
    scanf("%d%d", &a, &b);
    int gcd = ExGcd(a, b, x, y);
    printf("%d
", ((x%b)+b)%b);

    return 0;
}