bzoj [Noi2002]Savage 扩展欧几里得

枚举m，n^2判断

对于野人i,j，(H[i]+x*S[i])%m==(H[j]+x*S[j])%m,且x<=O[i]&&x<=O[j],他们才有可能相遇

化简得:(S[i]-S[j])*x+y*m=C[j]-C[i],扩欧解x最小值，判断

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define N 18
using namespace std;
int H[N],S[N],O[N],n,m,X,Y;
bool bo;
int exgcd(int a,int b,int &x,int &y){
	if(b==0){
		x=1;y=0;
		return a;
	}
	int gcd=exgcd(b,a%b,x,y);
	int t=x;
	x=y;
	y=t-(a/b)*x;
	return gcd;
}
bool judge(int x,int y,int mm){
	//printf("%d  %d  %d\n",x,y,mm);
	if(S[x]<S[y]) swap(x,y);
	int a=S[x]-S[y],b=mm,c=H[y]-H[x];
	int d=exgcd(a,b,X,Y);
	if(c%d!=0) return 0;
	X*=c/d;
	int bd=b/d;
	X=((X%bd)+bd)%bd;
	if(X<=O[x]&&X<=O[y]) return 1;
	return 0;
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d%d%d",&H[i],&S[i],&O[i]);
		m=max(m,H[i]); H[i]--;
	}
	for(;;m++){
		bo=0;
		for(int i=1;i<=n;i++){
			for(int j=i+1;j<=n;j++)
				if(judge(i,j,m)){bo=1;break;}
			if(bo==1)break;
		}if(bo==0)break;
	}
	printf("%d\n",m);
	return 0;
}