Describe：模拟大水题

　　code：

#include
     
      #define INF 214748364#define eps 1e-9#define rep1(a,b) for(register long long i=(a);i<=(b);i++)#define rep2(a,b) for(register long long j=(a);j<=(b);j++)using namespace std;struct JXCjulao{	long long x1,x2,y1,y2;}a[100110];long long n,x,y; inline long long read(){	long long ret=0,f=1;char ch=getchar();	while (ch<'0'||ch>'9') {if (ch=='-') f=-f;ch=getchar();}	while (ch>='0'&&ch<='9') ret=ret*10+ch-'0',ch=getchar();	return ret*f;}inline double read2(){    double X=0,Y=1.0;long long w=0;char ch=0;    while(!isdigit(ch)){w|=ch=='-';ch=getchar();}    while(isdigit(ch))X=X*10+(ch^48),ch=getchar();    ch=getchar();    while(isdigit(ch)) X+=(Y/=10)*(ch^48),ch=getchar();    return w?-X:X;}inline void write(long long x){	if(x<0){putchar('-');write(-x);return;}    if(x/10) write(x/10);putchar(x%10+'0');}int main(){	//freopen("carpet.in","r",stdin);    //freopen("carpet.out","w",stdout);    n=read();    for(int i=1;i<=n;i++)a[i].x1=read(),a[i].y1=read(),a[i].x2=read()+a[i].x1,a[i].y2=read()+a[i].y1;             //求地毯坐标    x=read(),y=read();for(int i=n;i>=1;i--)if(x<=a[i].x2&&x>=a[i].x1&&y<=a[i].y2&&y>=a[i].y1){cout<