POJ 1061 青蛙的约会 扩展欧几里得

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;

typedef long long ll;
ll exgcd(ll a, ll b, ll&x, ll&y) {
   if (b == 0) {
       x = 1;
       y = 0;
       return a;
   }
   ll r = exgcd(b, a%b, y, x);
   ll t = x;
   y = y - a/b*t;
   return r;
}
bool modular_linear_equation(ll a, ll b, ll n) {
    ll x, y, x0, i;
    ll d = exgcd(a, n, x, y);
    if (b%d)
    {
        printf("Impossible\n");
        return false;
    }
    x0 = x*(b/d)%n; //x0为方程的一个特解，可以为正也可以为负。题目要求的是最小的非负数
    ll ans;
    if(x0<0)
    {
        ans=x0;
        for(i = 0;ans<0; i++)
            ans=(x0 + i*(n/d))%n;
    }
    else if(x0>0)
    {
        ans=x0;
        ll temp;
        for(i=0;ans>=0;i++)
        {
            temp=ans;
            ans=(x0 - i*(n/d))%n;
        }
        ans=temp;
    }
    else
        ans=n; //此时x0=0，但是结果一定会大于0，所以要加一个n
    printf("%I64d\n",ans);
    return true;
}
int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    ll x,y,m,n,L;
    while(~scanf("%I64d%I64d%I64d%I64d%I64d",&x,&y,&m,&n,&L))
    {
        ll temp1=m-n,temp2=y-x;
        if(temp1<0)
        {
            temp1=-1*temp1;
            temp2=-1*temp2;
        }
        modular_linear_equation(temp1,temp2,L);
    }
    return 0;
}