欧几里得+扩展的欧几里得算法+线性同余方程+中国剩余定理


http://www.nowamagic.net/academy/detail/40110119
欧几里得算法:
欧几里德算法又称辗转相除法,用于计算两个整数a,b的最大公约数
1gcd(a, b) = gcd(b, a mod b)
扩展欧几里德算法
由朴素欧几里得算法我们可以得出扩展欧几里德算法:对于不全为 0 的非负整数 a、b,gcd(a,b)表示 a,b 的最大公约数,必然存在整数对 x,y ,使得 gcd(a,b)=ax+by。
就是给两个整数 a,b 必然存在一对整数 x,y 使得 ax + by = gcd(a,b),这个定理又叫贝祖定理。
void gcdEx(const int a,const int b,int& x,int& y){
    if (b){
//      gcdEx(b,a % b,x,y);
//      int t = x;
//      x = y;
//      y = t - (int)(a / b) * y;
        // 上面四句可化简为下面两句
        gcdEx(b,a % b,y,x);
        y -= (int)(a / b) * x;
    } else {
        x = 1;
        y = 0;
    }
}
http://www.nowamagic.net/academy/detail/40110122
已知等式 47x + 30y = 1; 求x,y的整数解。
设扩展欧几里德算法为gcdEx(a,b,x,y) 其中a,b为已知,x,y为所求。那么gcdEx(a,b,x,y)函数就有如下的计算过程:
解释一下这个过程:
1. 左边是利用朴素欧几里德算法相除得到一系列中间值,gcdEx(47,30,x,y),a = 47,b = 30,所以 gcd(47, 30) = gcd(b, a%b) = gcd(30, 17),直到b=0。
2. 当b=0时,可以得知x=1,y=0,当知道这两个值的时候,就可以开始根据扩展欧几里德算法倒推了。还记得那个定理吗?这里再说一次:
1x1 = y2
2y1 = x2 - (int)(a/b) * y2
3. 往上一步的x,y看作x1,y1,下面“旧”的x,y看作x2,y2,利用朴素欧几里德算法推导出的结论,x1=y2,y1=x2-(a/b)*y2。其中a,b就是当前gcdEx中的a,b.当这样递归回去到了第一个调用点时,所得到的x,y就是结果,即x=-7,y=11;
扩展欧几里得算法就是不断地的将b放小,直至b等于0,最后反推求回x和y。前面不断的递归求解一定会有个时候 b=0,所以递归可以结束。结束之后再利用定理反推。

http://www.nowamagic.net/academy/detail/40110128
小珂是一名初中生,她现在很苦恼,因为老师布置了一个让她苦恼的作业,你能不能帮助她呢?题目信息如下:
已知二元一次方程 a*x+b*y=n,判断这个二元一次方程有没有整数解,x、y为未知数,其中a、b、n都为整数且不等于零,同时满足0 < a,b,n < 2^16-1。
根据扩展欧几里德算法我们知道,对于与不完全为 0 的非负整数 a 和 b,gcd(a,b)表示 a 和 b 的最大公约数,那么存在唯一的整数 x、y,使得 gcd(a,b)=ax+by。
那么问题现在就简单了,就是比较 gcd(a,b) 与 n,如果他们模为0,那么就 a*x+b*y=n 就有解,否则无解。
扩展欧几里德算法的应用主要有以下三方面:
(1)求解不定方程;
(2)求解模线性方程(线性同余方程);
(3)求解模的逆元;


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