poj 1061 青蛙的约会（扩展欧几里得）

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http://poj.org/problem?id=1061

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。其中基础解x0 = x'*(c/d)%b(其中x'为a*x'+b*y' = gcd(a,b)的解)；通解为xi = x0 + i * (b/d)。

#include <stdio.h>#include <iostream>#include <algorithm>#include <set>#include <map>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#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;}

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