简介:欧几里德算法的应用
题目链接:poj 1061
解题思路:
A = m - n , B = y - x , N = L , 题目转化为求解模线性方程Ak ≡ B(mod N)
预处理:若m小于n,交换m与n,x与y
代码:
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
long long extend_gcd(long long a,long long b,long long &x,long long &y){
if(b==0){
x=1,y=0;
return a;
}
long long r=extend_gcd(b,a%b,y,x);
y-=a/b*x;
return r;
}
int main()
{
long long x,y,m,n,L,ans;
while(~scanf("%lld %lld %lld %lld %lld",&x,&y,&m,&n,&L))
{
if(m<n)
{
swap(x,y);
swap(m,n);
}
long long a=m-n,b=y-x,X,Y;
if(b<0) b+=L; //变成正数
long long d=extend_gcd(a,L,X,Y);
if(b%d==0)
{
X%=L,X+=L,X%=L;
ans=X*(b/d)%(L/d);
}
else
ans=-1;
if(ans==-1)
printf("Impossible\n");
else
printf("%lld\n",ans);
}
return 0;
}