Day6-T1
Describe:模拟大水题
code:
#include<bits/stdc++.h>
#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<<i;return 0;} //从后向前找符合条件的地毯
cout<<-1;return 0; //找不到puts(0);
}

浙公网安备 33010602011771号