poj1061 模線性方程

模板題

首先  x + k*m ≡ y + k* n (mod L) 轉化成 k(m - n) ≡ y - x(mod L)

套模板得到最小的解，但是要求是正的最小解，進行取模L

#include<iostream>
#include<cstdio>
using namespace std;
#define LL long long
inline  LL exgcd(LL a, LL b, LL &x, LL &y){/// ax + by = gcd(a, b);
    LL r;
    if(b){
        r =  exgcd(b, a%b, x, y);
        LL k = x;
        x = y;
        y = k - a/b*y;
    }
    else{
         y = (x = 1) - 1;
         return a;
    }
    return r;
}
LL mod_line(LL a, LL b, LL n)/// ak≡b(mod n) , ax + ny = b  x = x1*gcd(a, n)/d * i(1...gcd(a, n)); k最小爲 x1*gcd(a,n)/d
{
    LL x, y;
    LL d = exgcd(a, n, x, y);
    if(b%d != 0){
        return 0;
    }
    else{
        x = x*b/d;
        LL mod = n/d;
        if(mod < 0) mod = -mod;
        return (x % mod + mod) %mod;
    }
}
int main(){

    LL x, y, m, n, L;
    while(scanf("%lld%lld%lld%lld%lld", &x, &y, &m, &n, &L) != EOF){
        /// x + k*m = y + k*n  k*(m - n) = y - x
        LL a = (m - n);
        LL b = (y - x);
        LL k = mod_line(a, b, L);
        if(k == 0){
            printf("Impossible\n");
        }
        else{
            printf("%lld\n", k);
        }
    }
    return 0;
}