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;

}