POJ.1061 青蛙的约会 (拓展欧几里得)

POJ.1061 青蛙的约会 (拓展欧几里得)

题意分析

我们设两只小青蛙每只都跳了X次，由于他们相遇，可以得出他们同余，则有：

代码总览

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
void exgcd(ll a, ll b, ll& d, ll& x, ll &y)
{
    if(!b){
        d = a,x = 1,y = 0;
    }else{
        exgcd(b, a % b, d, y, x);
        y -= x * (a / b);
    }
}
int main()
{
    //freopen("in.txt","r",stdin);
    ll x,y,n,m,l;
    while(scanf("%I64d %I64d %I64d %I64d %I64d",&x,&y,&m,&n,&l)!= EOF){
        ll a = m-n, b = l, c = y-x, X, Y,gcd;
        exgcd(a,b,gcd,X,Y);
        if(c % gcd ){
            printf("Impossible
");
            continue;
        }else{
            ll t = fabs(b/gcd);
            X = X * (c / gcd);
            ll k = X * gcd / b;
            X = X - b / gcd * k;
            if(X <0) X+=t;
            printf("%I64d
",X);
        }
    }
    return 0;
}