BZOJ1477: 青蛙的约会

exgcd

( x + m*t ) - ( y + n*t ) = k * L  =>  ( n - m ) * t + k * L = x - y

题目链接

//Serene
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
using namespace std;
long long l,m,n,x,y;

void exgcd(long long a,long long b,long long &x,long long &y) {
    if(!b) {
        x=1;y=0;return ;
    }
    exgcd(b,a%b,x,y);
    swap(x,y);y=y-a/b*x;
}

long long gcd(long long a,long long b) {
    return b? gcd(b,a%b):a;
}

int main() {
    scanf("%lld%lld%lld%lld%lld",&x,&y,&m,&n,&l);
    long long a,b,c,gg;
    a=n-m;b=l;c=x-y;
    gg=gcd(a,b);
    if(c%gg) {
        printf("Impossible");
        return 0;
    }
    a/=gg;b/=gg;c/=gg;
    exgcd(a,b,x,y);
    x=((c*x)%b+b)%b;
    if(!x)x+=b;
    printf("%lld",x);
    return 0;
}