总结
自己同余方程理解的不深,exgcd更理解的不深,那还搞个屁,自己能写出来,我可以倒立拉稀加直播。
解析
建立两个方程组相减得
(m-n)X+L*Y=y-x;
ax+by=c
exgcd要保证a和b和c都是正整数。题目中a,b都可以为负数,y的个数我们不关心它,因为他是个孤儿 ,
然后我们可以对a和b求模为最小正整数,进行exgcd运行,维护x,就好。
题目链接
/*
____________ ______________ __
/ _________ /\ /_____ _____/\ / /\
/ /\ / / \\ / /\ \ \ / / \
/ / \_____/ / / \__/ / \____\/ / / /
/ / / / / / / / / / / /
/ / / / / / / / / / / /
/ / / / / / / / / / / /
/ /___/____/ / / / / / / /___/________
/____________/ / /__/ / /______________/\
\ \ / \ \ / \ \ \
\____________\/ \__\/ \______________\/
___ ___ ___ __________
/ /\ / /\ / /\ /_______ /\
/ /__\___/ / \ / / \ \ / / \
/____ ____/ / / / / \____/ / /
\ / /\ \ / / / / / / /
\_/ / \___\/ ___ / / / / / /
/ / / / /\ / / / / / /
/ / / / /__\__/ / / / /___/____
/___/ / /___________/ / /___________/\
\ \ / \ \ / \ \ \
\___\/ \___________\/ \___________\/
A CODE OF CBOY
*/
#include<bits/stdc++.h>
//typedef long long ll;
//#define ull unsigned long long
#define int long long
#define F first
#define S second
#define endl "\n"//<<flush
#define eps 1e-6
#define lowbit(x) (x&(-x))
#define PI acos(-1.0)
#define inf 0x3f3f3f3f
#define MAXN 0x7fffffff
#define INF 0x3f3f3f3f3f3f3f3f
#define pa pair<int,int>
#define ferma(a,b) pow(a,b-2)
#define pb push_back
#define all(x) x.begin(),x.end()
#define memset(a,b) memset(a,b,sizeof(a));
#define IOS ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);
using namespace std;
void file()
{
#ifdef ONLINE_JUDGE
#else
freopen("cin.txt","r",stdin);
// freopen("cout.txt","w",stdout);
#endif
}
int f(int a,int l)
{
if(a<=0)
{
int cnt=a/l;
if(a%l)
cnt++;
a+=cnt*l;
}
if(a==0)
a+=l;
return a;
}
int exgcd(int a,int b,int &x,int &y)
{
if(b==0)
{
x=1,y=0;
return a;
}
int x1,y1;
int gcd=exgcd(b,a%b,x1,y1);
x=y1,y=x1-a/b*y1;
return gcd;
}
signed main()
{
IOS;
//file();
int x,y,m,n,l;
cin>>x>>y>>m>>n>>l;
int a=m-n,b=l,c=y-x;
a=f(a,l);
c=f(c,l);
int gcd=exgcd(a,b,x,y);
if(c%gcd)
{
cout<<"Impossible"<<endl;
return 0;
}
int k1=b/gcd;
x=(x%k1+k1)%k1;
cout<<c/gcd*x%k1<<endl;
return 0;
}