？？？至于那么麻烦吗？？？
```cpp
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
using namespace std;
int n,x,y,ans=-1;
int a[10005],b[10005],g[10005],k[10005];
int main()
{
cin >> n;
for(int i=1;i<=n;i++)
cin >> a[i] >> b[i] >> g[i] >> k[i];
cin >> x >> y;
for(int i=1;i<=n;i++)
if(x>=a[i]&&a[i]+g[i]>=x&&y>=b[i]&&y<=b[i]+k[i])
ans = i;
cout << ans;
}
```
by AVALON_7 @ 2017-10-25 12:55:23
等等等等你那个不叫反转了叫重新赋值了
by AVALON_7 @ 2017-10-25 14:05:48
```cpp
#include<stdio.h>
int main()
{
int a[10000],d[10000],f[10000];
int b[10000],c[10000],e[10000];
int n,x,y,i,m=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d%d%d%d",&a[i],&b[i],&c[i],&d[i]);
e[i]=a[i]+c[i];//同上
f[i]=b[i]+d[i];//同上
}
scanf("%d%d",&x,&y);
for(i=0;i<n;i++)
{
int i1=n-1-i;
if(x>=a[i1] && x<=e[i1] &&y>=b[i1]&&y<=f[i1])//同上
{
printf("%d",n-i);//如果正确 打印结果
return 0;
}
}
printf("%d",-1);return 0;
}
```
by AVALON_7 @ 2017-10-25 14:09:40
~~呵呵~~
by 蒻得不行 @ 2017-10-26 16:13:31
呵呵 O(∩\_∩)O~
by 洪荒无极 @ 2017-11-17 22:51:37