BZOJ 1407 exgcd

枚举 山洞个数，每次n^2  枚举两个野人之间是否会发生冲突.

  相当于求p[i]*x+c[i] = p[j]*x+c[j] mod m (天数)

    ----------> (p[i]-p[j])*x-m*y = c[j]-c[i] .. exgcd 求解.

#include <iostream>#include <cstring>#include <algorithm>#include <stdio.h>#define MAXN 17using std::max;int n,p[MAXN],c[MAXN],l[MAXN],mx;  template<typename _t>inline _t read(){    _t x=0,f=1;    char ch=getchar();    for(;ch>'9'||ch<'0';ch=getchar())if(ch=='-')f=-f;    for(;ch>='0'&&ch<='9';ch=getchar())x=x*10+(ch^48);    return x*f;} inline int gcd(int x,int y){return y==0?x:gcd(y,x%y);} void gcd(int a,int b,int &x,int &y){if(b==0){x=1;y=0;return;}gcd(b,a%b,x,y);int t=x;x=y;y=t-a/b*y;} inline bool Judge(int tot){    // (p[i]-p[j])*x -my = c[j]-c[i];    for(register int i=1;i<=n;i++){        for(register int j=i+1;j<=n;j++){            int A,B,C,x,y;            A = p[i]-p[j],B=c[j]-c[i],C=tot;            int t = gcd(A,C);            if(B%t==0){                A/=t;B/=t;C/=t;                gcd(A,C,x,y);                if(C<0)C=-C;                x = ((B*x)%C+C)%C;                if(!x)x+=C;                if(x<=l[i]&&x<=l[j])return 0;            }        }    }    return 1;} int main(){    n=read<int>();    for(int i=1;i<=n;i++){        c[i]=read<int>(),        p[i]=read<int>(),        l[i]=read<int>();        mx=max(mx,c[i]);    }    while(!Judge(mx++));mx--;    printf("%d\n",mx);}