poj 1061 青蛙的约会（扩展欧几里得）-白红宇

poj 1061 青蛙的约会（扩展欧几里得）

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发布时间：2019-06-18

本文共 1173 字，大约阅读时间需要 3 分钟。

思路：设它们跳了t次相遇，那么有 (x+t*m)-(y+t*n) = z*l(z是一个整数，表示它们路程差是l的z倍)，变形得

(n-m)*t + z*l = (x-y);

令 a = n-m; b = l; c = x-y;

那么原式变为 a*t + z*b = c;

扩展欧几里得模板，求解形如a*x + b*y = gcd(a,b)方程。

LL extend_gcd(LL a, LL b, LL &x, LL &y){	if(b == 0)	{		x = 1;		y = 0;		return a;	}	LL d = extend_gcd(b,a%b,x,y);	LL t = x;	x = y;	y = t-a/b*y;	return d;}

方程a*x + b*y = c有解的前提是 gcd(a,b) | c,在这个基础上方程有d=gcd(a,b)个不同的解。当中基础解x0 = x'*(c/d)%b(当中x'为a*x'+b*y' = gcd(a,b)的解)；通解为xi = x0 + i * (b/d)。

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                #define LL long long#define _LL __int64#define eps 1e-8using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 10;LL extend_gcd(LL a, LL b, LL &x, LL &y){ if(b == 0) { x = 1; y = 0; return a; } LL d = extend_gcd(b,a%b,x,y); LL t = x; x = y; y = t-a/b*y; return d;}int main(){ LL x,y,m,n,l; LL a,b,c,d; while(~scanf("%lld %lld %lld %lld %lld",&x,&y,&m,&n,&l)) { a = n-m; b = l; c = x-y; LL t,z; d = extend_gcd(a,b,t,z); if(c%d != 0) { printf("Impossible\n"); continue; } t = t*(c/d); t = (t%b+b)%b; printf("%lld\n",t); } return 0;}