You can not select more than 25 topics
Topics must start with a letter or number, can include dashes ('-') and can be up to 35 characters long.
|
|
|
|
|
#include <bits/stdc++.h>
|
|
|
|
|
|
|
|
|
|
|
|
using namespace std;
|
|
|
|
|
|
typedef long long LL;
|
|
|
|
|
|
//扩展欧几里得
|
|
|
|
|
|
LL exgcd(LL a, LL b, LL &x, LL &y) {
|
|
|
|
|
|
if (!b) {
|
|
|
|
|
|
x = 1, y = 0;
|
|
|
|
|
|
return a;
|
|
|
|
|
|
}
|
|
|
|
|
|
LL d = exgcd(b, a % b, y, x);
|
|
|
|
|
|
y -= a / b * x;
|
|
|
|
|
|
return d;
|
|
|
|
|
|
}
|
|
|
|
|
|
int main() {
|
|
|
|
|
|
/* 原始方程 4x+6y=10 求x,y的一组解
|
|
|
|
|
|
|
|
|
|
|
|
分析及求解步骤：
|
|
|
|
|
|
此题中a=4,b=6,c=10
|
|
|
|
|
|
1. 求d=gcd(a,b)=gcd(4,6)=2
|
|
|
|
|
|
2. 判断 2|c即 10%2=0，方程有解，如果不能整除，此处可以直接返回方程无整数解
|
|
|
|
|
|
3. 化简方程为 a/d*x+b/d*x=c/d 得到 4/2*x+6/2*y=10/2
|
|
|
|
|
|
=> 2x+3y=5 此方程与原始方程解是一样的，我们最终对这个方程下手。
|
|
|
|
|
|
4. 如可解此方程呢？我们有一个好用的定理：扩展欧几里得可以解类似的方程
|
|
|
|
|
|
ax+by=gcd(a,b)!
|
|
|
|
|
|
但问题是按扩欧的描述，我们现在能计算的是 2x+3y=1的方程解！！！
|
|
|
|
|
|
不是原方程2x+3y=5的解!!
|
|
|
|
|
|
该怎么办呢？
|
|
|
|
|
|
设 x0,y0为2x+3y=1的解，那么原方程的一组解就是：
|
|
|
|
|
|
x1=5*x0,y1=5*y0!!!!
|
|
|
|
|
|
|
|
|
|
|
|
为什么呢？原因很简单： 2x+3y=1 左右都乘以5
|
|
|
|
|
|
2*(x*5)+3*(y*5)=5
|
|
|
|
|
|
对比一下原方程 x1=x0*5,y1=y0*5啊~
|
|
|
|
|
|
*/
|
|
|
|
|
|
LL x, y;
|
|
|
|
|
|
exgcd(2, 3, x, y);
|
|
|
|
|
|
printf("2x+3y=1 -> x=%lld y=%lld\n", x, y);
|
|
|
|
|
|
printf("2x+3y=5 -> x=%lld y=%lld", 5 * x, 5 * y);
|
|
|
|
|
|
return 0;
|
|
|
|
|
|
}
|
