测试数据 #0: Accepted, time = 0 ms, mem = 888 KiB, score = 10
测试数据 #1: Accepted, time = 15 ms, mem = 888 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 888 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 888 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 892 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 888 KiB, score = 10
测试数据 #6: Accepted, time = 15 ms, mem = 888 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 888 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 888 KiB, score = 10
测试数据 #9: Accepted, time = 15 ms, mem = 884 KiB, score = 10
Accepted, time = 75 ms, mem = 892 KiB, score = 100
代码
type aa=record
a,b,g,k:longint;
end;
var
x,y,i,n:longint;
kk:array[0..10000]of aa;

begin

readln(n);
for i:=1 to n do readln(kk[i].a,kk[i].b,kk[i].g,kk[i].k);
readln(x,y);
for i:=n downto 1 do
begin
if (x>=kk[i].a) and (y>=kk[i].b) and (x<=kk[i].a+kk[i].g-1) and (y<=kk[i].b+kk[i].k-1) then begin writeln(i);readln; halt; end;
end;

writeln('-1');
readln;

end.

模拟。数组从后往前扫，找到直接输出就行。

Pascal Code

var
a,b,g,k:array[1..10000] of longint;
n,x,y,i:longint;
function iscover(a,b,g,k:longint):boolean; //判断是否覆盖的函数
begin
if x<a then exit(false);
if y<b then exit(false);
if x>a+g then exit(false);
if y>b+k then exit(false);
exit(true);
end;
begin
read(n);
for i:=1 to n do
readln(a[i],b[i],g[i],k[i]);
readln(x,y);
for i:=n downto 1 do
if iscover(a[i],b[i],g[i],k[i])
then begin
writeln(i);
halt;
end;
writeln(-1);
end.

编译成功

测试数据 #0: Accepted, time = 15 ms, mem = 980 KiB, score = 10

测试数据 #1: Accepted, time = 0 ms, mem = 984 KiB, score = 10

测试数据 #2: Accepted, time = 0 ms, mem = 976 KiB, score = 10

测试数据 #3: Accepted, time = 0 ms, mem = 980 KiB, score = 10

测试数据 #4: Accepted, time = 15 ms, mem = 980 KiB, score = 10

测试数据 #5: Accepted, time = 0 ms, mem = 984 KiB, score = 10

测试数据 #6: Accepted, time = 15 ms, mem = 976 KiB, score = 10

测试数据 #7: Accepted, time = 7 ms, mem = 976 KiB, score = 10

测试数据 #8: Accepted, time = 15 ms, mem = 976 KiB, score = 10

测试数据 #9: Accepted, time = 15 ms, mem = 976 KiB, score = 10

Accepted, time = 82 ms, mem = 984 KiB, score = 100

评测结果
编译成功

Free Pascal Compiler version 2.6.2 [2013/02/12] for i386
Copyright (c) 1993-2012 by Florian Klaempfl and others
Target OS: Win32 for i386
Compiling foo.pas
foo.pas(6,3) Note: Local variable "i1" not used
foo.pas(6,6) Note: Local variable "j1" not used
foo.pas(6,15) Note: Local variable "a" not used
foo.pas(6,17) Note: Local variable "b" not used
foo.pas(6,19) Note: Local variable "g" not used
foo.pas(6,21) Note: Local variable "k" not used
Linking foo.exe
22 lines compiled, 0.1 sec , 28208 bytes code, 1628 bytes data
6 note(s) issued
测试数据 #0: Accepted, time = 15 ms, mem = 772 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 772 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 772 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 772 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 772 KiB, score = 10
测试数据 #5: Accepted, time = 15 ms, mem = 772 KiB, score = 10
测试数据 #6: Accepted, time = 15 ms, mem = 772 KiB, score = 10
测试数据 #7: Accepted, time = 15 ms, mem = 772 KiB, score = 10
测试数据 #8: Accepted, time = 15 ms, mem = 772 KiB, score = 10
测试数据 #9: Accepted, time = 15 ms, mem = 772 KiB, score = 10
Accepted, time = 90 ms, mem = 772 KiB, score = 100
#代码

type AA=record
a,b,g,k:longint;
end;
var
i,i1,j1,x,y,n,a,b,g,k,p:longint;
f:array[0..10000]of AA;
begin
readln(n);
for i:=1 to n do readln(f[i].a,f[i].b,f[i].g,f[i].k);
readln(x,y);
i:=n;
p:=-1;
while(i>0) do begin
if (f[i].a+f[i].g-1>=x)and(x>=f[i].a)and(f[i].b+f[i].k-1>=y)and(y>=f[i].b) then
begin
p:=i;
break;
end;
dec(i);
end;
writeln(p);
end.

AC代码.....优化到最优...
var f:array[1..10000,1..4]of longint;
n,s,x,y,a:longint;
begin
readln(n); a:=0;
for s:=1 to n do
readln(f[s,1],f[s,2],f[s,3],f[s,4]);
readln(x,y);
for s:=1 to n do
if (x>=f[s,1])and(x<=f[s,3]+f[s,1])and(y>=f[s,2])and(y<=f[s,4]+f[s,2])
then a:=s;
if a<>0 then writeln(a) else writeln(-1);
end.

#include <cstdio>

const int MAXN = 10000+10;

int n, x, y, a[MAXN], b[MAXN], g[MAXN], k[MAXN];

inline int Solve(){
for (int i=n; i>0; i--)
if (a[i]<=x && x<=a[i]+g[i] && b[i]<=y && y<=b[i]+k[i]) return i;
return -1;
}

int main(){
scanf("%d", &n);
for (int i=1; i<=n; i++)
scanf("%d %d %d %d", &a[i], &b[i], &g[i], &k[i]);
scanf("%d %d", &x, &y);
printf("%d\n", Solve());
}

#include <cstdio>
int main()
{ int n;
scanf("%d", &n);
int a[n][4];
for (int i=0;i<n;i++)
for (int j=0;j<4;j++)
scanf("%d", &a[i][j]);
int x,y;
scanf("%d%d", &x, &y);
for (int i=n-1;i>=0;i--)
{ int xrange=a[i][0]+a[i][2],yrange=a[i][1]+a[i][3];
if ((x>=a[i][0] && x<=xrange) && (y>=a[i][1] && y<=yrange))
{ printf("%d\n", i+1); break;}
if (i==0)
printf("-1\n");
}

return 0;
}

#include "cstdio"
#include "vector"
using namespace std;
struct ditan{
int a;int b;int g;int k;
};
vector<ditan>v;
int n,x,y,ans=-1;
int main(){
scanf("%d\n",&n);
for (int i=0; i<n; i++) {
ditan tmp;
scanf("%d %d %d %d\n",&tmp.a,&tmp.b,&tmp.g,&tmp.k);
v.push_back(tmp);
}
scanf("%d %d\n",&x,&y);
for (int i=0; i<n; i++) {
if((v[i].a<=x&&x<v[i].a+v[i].g)&&(v[i].b<=y&&y<v[i].b+v[i].k))ans=i+1;
else;
}
if(ans>-1)printf("%d\n",ans);
else printf("-1\n");
}

Vijos 题解：http://hi.baidu.com/umule/item/2c997f8ed9600fdae596e017
有疑问请留言 共同进步

var
n,x,y,i,j,k,a,b,g,fx,fy:longint;
f:array[1..100000,1..6] of longint;
begin
readln(n);
for i:=1 to n do
begin
read(a,b,g,k);
f[i,1]:=a;f[i,2]:=b;f[i,3]:=g;f[i,4]:=k;
f[i,5]:=a+g;f[i,6]:=b+k;
end;
readln;
read(x,y);
j:=-1;
for i:=1 to n do
if (f[i,1]>x)or(f[i,2]>y) then continue
else
begin
if (f[i,5]>=x)and(f[i,6]>=y) then j:=i;
end;
writeln(j);
end.

有点水。

(￣_￣|||) 竟然把复制的测试数据提交了 WA了一次

Var x,y,x1,y1:array[1..10000] of longint;
a,b,a1,b1,n,i,j:longint;
Begin
readln(n);
For i:=1 to n do
Begin
readln(a,b,a1,b1);
a1:=a+a1; b1:=b+b1;
x[i]:=a; y[i]:=b;
x1[i]:=a1; y1[i]:=b1;
End;
Readln(a,b);
For i:=n downto 1 do
If (a>=x[i]) and (a<=x1[i]) and (b>=y[i]) and (b<=y1[i]) then
begin
writeln(i);
halt;
end;
writeln(-1);
readln;
end.

#include <cstdio>

using namespace std;

#define MAXN 10000

int q[MAXN][4];
int n;
int x,y;

int main(){
scanf("%d",&n);
for (int i=0;i<n;i++){
scanf("%d%d%d%d",&q[i][0],&q[i][1],&q[i][2],&q[i][3]);
}
scanf("%d%d",&x,&y);
int ans=-1;
for (int i=0;i<n;i++){
if (x>=q[i][0]&&x<=q[i][0]+q[i][2]&&y>=q[i][1]&&y<=q[i][1]+q[i][3]){
ans=i+1;
}
}
printf("%d\n",ans);
return 0;
}

#include

using namespace std;

int main()

{

int m,n,t,i,s=-1;

int x[10000],y[10000],a[10000],b[10000];

cin>>t;

for(i=0;i>x[i]>>y[i]>>a[i]>>b[i];

cin>>m>>n;

for(i=t-1;i>=0;i--)

if(m=x[i] && n=y[i]) {s=i+1; break;}

cout

测评机又抽啦！！！！

本人提交：第一次，超时一个点；第二次，超时一个点；第三次，超时一个点，第四次，超时三个点；第五次，AC!!!!!!!!!!!!!!!

同样的程序啊！

不过话说回来，这题一点也不难哈，18行搞定！

program P1736;

var

a,b,c,d:array[1..10000] of longint;

g,k,x,y:longint;

i,n:integer;

begin

readln(n);

for i:=1 to n do

begin

read(a[i],b[i],g,k);

c[i]:=a[i]+g;

d[i]:=b[i]+k;

end;

read(x,y);

for i:=n downto 1 do

if (x>=a[i])and(x=b[i])and(y

编译通过...

├ 测试数据 01：答案正确... (0ms, 736KB)

├ 测试数据 02：答案正确... (0ms, 736KB)

├ 测试数据 03：答案正确... (0ms, 736KB)

├ 测试数据 04：答案正确... (0ms, 736KB)

├ 测试数据 05：答案正确... (0ms, 736KB)

├ 测试数据 06：答案正确... (0ms, 736KB)

├ 测试数据 07：答案正确... (0ms, 736KB)

├ 测试数据 08：答案正确... (0ms, 736KB)

├ 测试数据 09：答案正确... (0ms, 736KB)

├ 测试数据 10：答案正确... (0ms, 736KB)

---|---|---|---|---|---|---|---|-

Accepted / 100 / 0ms / 736KB

SecretAgent的思路是对的

二维线段树900ms+...

沙发~

把坐标按顺序存好

反向检查输出答案

//读入优化

#include<cstdio>
using namespace std;
template<class T> inline void read(T &_a){
    bool f=0;int _ch=getchar();_a=0;
    while(_ch<'0' || _ch>'9'){if(_ch=='-')f=1;_ch=getchar();}
    while(_ch>='0' && _ch<='9'){_a=(_a<<1)+(_a<<3)+_ch-'0';_ch=getchar();}
    if(f)_a=-_a;
}
struct ditan{ int x,y,g,k; } t[10001];
int n,x,y,ans;
int main()
{
    read(n);
    for (int i=1;i<=n;i++)
     read(t[i].x),read(t[i].y),read(t[i].g),read(t[i].k);
    read(x);read(y);
    for (int i=1;i<=n;i++)
     if(x>=t[i].x&&x<=t[i].x+t[i].g&&y>=t[i].y&&y<=t[i].y+t[i].k)
      ans=i;
    printf("%d",ans);
    return 0;
}

几分钟就解决了

简单的模拟题吧。。。