~~标题可能爆炸了不要介意~~
by lizer @ 2018-11-18 22:38:22
来几个dalao指教指教啊...
by lizer @ 2018-11-18 22:38:48
```cpp
for(i=n;i>=1;i--)
if(x>=a[i] && x<=a[i]+c[i] && y>=b[i] && y<=b[i]+d[i]){
printf("%d",i);
return 0;
}
```
这样（变量名不一样不要介意）
by 梧桐灯 @ 2018-11-18 22:39:53
@[光随影走](/space/show?uid=31193) emmm不~~是很~~明白...
by lizer @ 2018-11-18 22:44:56
@[liyize](/space/show?uid=111010) 就是判断一个地毯是否覆盖一个坐标只需O(1)的复杂度
by 梧桐灯 @ 2018-11-18 22:46:27
@[光随影走](/space/show?uid=31193) emmm~~好像明白了~~
我去试试
by lizer @ 2018-11-18 22:49:13
@[光随影走](/space/show?uid=31193) A辣
蟹蟹神犇
```cpp
#include <iostream>
using namespace std;
struct carpet{
int a;
int b;
int g;
int k;
}a[100001];
int main()
{
int n,x,y;
cin>>n;
for(int i=1;i<=n;i++) cin>>a[i].a>>a[i].b>>a[i].g>>a[i].k;
cin>>x>>y;
for(int i=n;i>=1;i--)
if(x>=a[i].a&& x<=a[i].a+a[i].g&&y>=a[i].b&&y<=a[i].b+a[i].k){
cout<<i;
return 0;
}
return 0;
}
```
代码也简洁了好多呢
by lizer @ 2018-11-18 22:54:48