青蛙的旅行 poj 1061

// 参考->青蛙的约会 exgcd解同余方程

定理证明->点我

/*
 * @Promlem: 
 * @Time Limit: ms
 * @Memory Limit: k
 * @Author: pupil-XJ
 * @Date: 2019-10-20 17:19:56
 * @LastEditTime: 2019-10-20 18:52:09
 */
#include<cstdio>
#include<cstring>
#include<cmath>
#include<iostream>
#include<string>
#include<algorithm>
#include<vector>
#include<queue>
#include<stack>
#include<set>
#include<map>
#define rep(i, n) for(int i=0;i!=n;++i)
#define per(i, n) for(int i=n-1;i>=0;--i)
#define Rep(i, sta, n) for(int i=sta;i!=n;++i)
#define rep1(i, n) for(int i=1;i<=n;++i)
#define per1(i, n) for(int i=n;i>=1;--i)
#define Rep1(i, sta, n) for(int i=sta;i<=n;++i)
#define L k<<1
#define R k<<1|1
#define mid (tree[k].l+tree[k].r)>>1
using namespace std;
const int INF = 0x3f3f3f3f;
typedef long long ll;

inline ll gcd(ll x, ll y) { return (!y) ? x : gcd(y, x%y); }

inline ll exgcd(ll a, ll b, ll &x, ll &y) {
    if(!b) {
        x = 1;
        y = 0;
        return a;
    }
    ll ans = exgcd(b, a%b, x, y);
    ll temp = x;
    x = y;
    y = temp - a/b*y;
    return ans;
}

int main() {
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    ll x, y, m, n, l;
    cin >> x >> y >> m >> n >> l;
    ll a = l, b = n-m, c = x-y, d = gcd(a, b);
    if(c % d != 0) cout << "Impossible
";
    else {
        exgcd(a, b, x, y);
        cout << ((y*c/d)%(l/d)+(l/d))%(l/d) << "
";
    }
    return 0;
}