5

#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<algorithm>

using namespace std;

int main()
{
int n,x,y,ans=-1;
int a[100005],b[100005],g[100005],k[100005];
cin>>n;
for(int i=1;i<=n;i++)
{
cin>>a[i]>>b[i]>>g[i]>>k[i];
}
cin>>x>>y;
for(int j=1;j<=n;j++)
{
if(x>=a[j]&&x<=a[j]+g[j]&&y>=b[j]&&y<=b[j]+k[j])
{
ans=j;
}
}
cout<<ans;
return 0;
}

吴哥——传奇 @ 7 年前

太复杂了，还是我的好
海之精华 @ 6 年前

判断地毯可以从n到1，因为地毯是最后的在最上面，判断成功直接输出就行了

2

Rssqzqyp LV 7 @ 7 年前

好水的题啊啊啊啊啊啊啊啊啊
一开始还以为很难
然后发现n这么小随随便便就过了
```cpp
#include<bits/stdc++.h>
using namespace std;
int n;
int a[1000000][4];
int main()
{
cin>>n;
for(int i=1;i<=n;i++)
cin>>a[i][0]>>a[i][1]>>a[i][2]>>a[i][3];
int t1,t2;
cin>>t1>>t2;
for(int i=n;i>=1;i--)

if(t1>=a[i][0]&&t1<=a[i][0]+a[i][2]&&t2>=a[i][1]&&t2<=a[i][1]+a[i][3])
{
cout<<i;
return 0;
}
cout<<-1;
return 0;

}
```

1

aph。 (chenqianrong) LV 10 @ 3 年前

#include<cstdio>
#include<iostream>
using namespace std;

int main()
{
    int i,a[10001][4],n,x,y,f=-1;
    cin>>n;
    for( i=1; i<=n; i++){
        cin>>a[i][0]>>a[i][1]>>a[i][2]>>a[i][3];
        a[i][2]+=a[i][0];
        a[i][3]+=a[i][1];
    }
    cin>>x>>y;
    for(i=n;i>=1;i--){
        if(a[i][0]<=x && a[i][1]<=y && a[i][2]>=x && a[i][3]>=y){
            cout<<i;
            f=1;
            break;
        }
    }
    if(f==-1)
        cout<<f;
    return 0;
}

1

血色X战魂 LV 8 @ 6 年前

Accepted

状态耗时内存占用

#1 Accepted 1ms 448.0 KiB
#2 Accepted 1ms 448.0 KiB
#3 Accepted 1ms 448.0 KiB
#4 Accepted 1ms 448.0 KiB
#5 Accepted 1ms 448.0 KiB
#6 Accepted 1ms 448.0 KiB
#7 Accepted 3ms 448.0 KiB
#8 Accepted 4ms 448.0 KiB
#9 Accepted 7ms 448.0 KiB
#10 Accepted 4ms 448.0 KiB

#include<iostream>
#include<algorithm>
using namespace std;
//呵了个呵
//大神万岁
//分割线--------------------------------------------------$
//结构定义区
struct ditan{
    int a,b,g,k;
    ditan()
    {
        a=b=g=k=0;
    }
}; 
//全局变量区
ditan dt[10001];
int n,x,y,ans=-1;
//函数声明区

//主函数开始！
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>dt[i].a>>dt[i].b>>dt[i].g>>dt[i].k;
     } 
    cin>>x>>y;
    for(int i=1;i<=n;i++)
    {
        if(x>=dt[i].a && y>=dt[i].b && x<=dt[i].a+dt[i].g && y<=dt[i].b+dt[i].k) ans=i;
    }
    cout<<ans;
    return 0;
}
//函数实现区

1

清风breeze LV 8 @ 7 年前

#include<cstdio>
struct carpet{
    int num;
    int a;
    int b;
    int g;
    int k;
}ca[10010];
int ans=-1;
int main(){
    int n, x, y;
    scanf("%d", &n);
    for(int i=1; i<=n; i++) scanf("%d%d%d%d", &ca[i].a, &ca[i].b, &ca[i].g, &ca[i].k);
    scanf("%d%d", &x, &y);
    for(int i=1; i<=n; i++){
        int _x=ca[i].a+ca[i].g;
        int _y=ca[i].b+ca[i].k;
        if(x<=_x&&x>=ca[i].a&&y<=_y&&y>=ca[i].b) ans=i;
    }
    printf("%d", ans);
    return 0;
}

0

季傲然 (aaron07) LV 7 @ 2 年前

#include<bits/stdc++.h>
using namespace std;
int main()
{
    int n,x,y;
    int a[10001],b[10001],g[10001],k[10001];
    int i;
    cin>>n;
    for(i=1;i<=n;i++)
    {
        cin>>a[i]>>b[i];
        cin>>g[i]>>k[i];
    }
    cin>>x>>y;
    for(i=n;i>=1;i--)
    {
        if((x>=a[i]&&x<=a[i]+g[i]) && (y>=b[i]&&y<=b[i]+k[i]))
        {
            cout<<i<<endl;
            return 0;
        }
    }
    cout<<-1<<endl;
    return 0;
}

0

季傲然 (aaron07) LV 7 @ 2 年前

#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,x,y;
int a[10001],b[10001],g[10001],k[10001];
int i;
cin>>n;
for(i=1;i<=n;i++)
{
cin>>a[i]>>b[i];
cin>>g[i]>>k[i];
}
cin>>x>>y;
for(i=n;i>=1;i--)
{
if((x>=a[i]&&x<=a[i]+g[i]) && (y>=b[i]&&y<=b[i]+k[i]))
{
cout<<i<<endl;
return 0;
}
}
cout<<-1<<endl;
return 0;
}

0

Jong-oo LV 8 @ 5 年前

#include <iostream>         //[2011提高组Day1-A]铺地毯
#include <algorithm>
using namespace std;
const int MAXN = 10002;

struct Node{
    int a, b;
    int g, k;
} F[MAXN];

bool Ar(int x, int y, int i)
{
    if(x >= F[i].a && x <= F[i].a + F[i].g)
        if(y >= F[i].b && y <= F[i].b + F[i].k)
            return true;
    return false;
}

int main()
{
    int n, a, b, g, k, x, y;
    cin >> n;
    for (int i = 1; i <= n; i++)
        cin >> F[i].a >> F[i].b >> F[i].g >> F[i].k;
    cin >> x >> y;
    int ans;
    for (ans = n; ans > 0; ans--)
        if(Ar(x, y, ans))
            break;
    if(ans)
        cout << ans << endl;
    else
        cout << -1 << endl;

    system("pause");
    return 0;

}

0

海之精华 LV 7 @ 6 年前

因为这里的n有点小（貌似正常是100000？）所以开个二维应该（？）可以……
代码是空间多时间短，如果要空间少的话可以拿一个地毯判一个
#include<iostream>
#include<cstdio>

using namespace std;

#define l long

const l N=100010;

l n;
l a[N],b[N],g[N],k[N];
l x,y;

l read()
{
l d=0,f=1;
char c=getchar();
while(c<'0' || c>'9'){if(c=='-')f=-1;c=getchar();}
while(c>='0' && c<='9'){d=d*10+c-'0';c=getchar();}
return d*f;
}

int main()
{
//freopen("carpet.in","r",stdin);
//freopen("carpet.out","w",stdout);
n=read();
for(l i=1;i<=n;i++){a[i]=read();b[i]=read();g[i]=read();k[i]=read();}
x=read();y=read();

for(l i=n;i>=1;i--)
if(x>=a[i] && x<=a[i]+g[i] && y>=b[i] && y<=b[i]+k[i])
{printf("%ld\n",i);return 0;}

printf("-1\n");return 0;
}

0

KotoriMinami LV 7 @ 6 年前

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin>>n;
    int a[4][n],i;
    for(i=0;i<n;i++)
        cin>>a[0][i]>>a[1][i]>>a[2][i]>>a[3][i];
    int x,y;
    cin>>x>>y;
    int total=0;
    for(i=0;i<n;i++)
        if(a[0][i]<=x&&a[1][i]<=y&&a[0][i]+a[2][i]>=x&&a[1][i]+a[3][i]>=y) total=i+1;
    if(total>0) cout<<total<<endl;
    else if(total==0) cout<<"-1"<<endl;
    return 0;
}

0

xianggulinger LV 8 @ 7 年前

......

#include <iostream>
using namespace std;
#define For(aHJEfaks, fwbjkWK, AENGIBv) for (int aHJEfaks = fwbjkWK; aHJEfaks <= AENGIBv; ++aHJEfaks)
#define For2(aHJEfaks, fwbjkWK, AENGIBv) for (auto aHJEfaks = fwbjkWK; aHJEfaks != AENGIBv; ++aHJEfaks)
int n;
int x[20000], y[20000], lx[20000], ly[20000];
int main(){
    cin >> n;
    For(i, 1, n){
        cin >> x[i] >> y[i] >> lx[i] >> ly[i];
    }
    int x0, y0;
    cin >> x0 >> y0;
    for (int j = n; j > 0; --j) {
        if (x[j] <= x0 && x0 <= x[j] + lx[j] && y[j] <= y0 && y0 <= y[j] + ly[j]){
            cout << j << endl;
            return 0;
        }
    }
    cout << -1 << endl;
    return 0;
}

0

Gadfly-Km LV 8 @ 7 年前

开二维爆炸orz...

#include<iostream>
const int maxn=10005;
int a[maxn],b[maxn],g[maxn],k[maxn];
using namespace std;
int main()
{
   int n,x,y,ans=-1;//-1方便输出 
   cin>>n;
   for (int i=1;i<=n;i++) cin>>a[i]>>b[i]>>g[i]>>k[i];//读入 
   cin>>x>>y;
   for (int i=1;i<=n;i++)
     if(a[i]<=x&&b[i]<=y&&a[i]+g[i]>=x&&b[i]+k[i]>=y) ans=i;/*地毯要覆盖一点，即左下方点必须
    小于指定点的横纵坐标（自己去试试），地毯长度必须大于指定点的横纵坐标 */
    cout<<ans<<endl;
    return 0;
}

0

李卓远 LV 7 @ 7 年前

#include<iostream>
using namespace std ;
struct ha{
int a,b,g,k,a1,b1;
}d[10000];
int main()
{
int i,n,mm,nn;
freopen("carpet.in","r",stdin);
freopen("carpet.out","w",stdout);
cin>>n;
for(i=1;i<=n;i++)
{
cin>>d[i].a>>d[i].b>>d[i].g>>d[i].k;
d[i].a1=d[i].a+d[i].g;
d[i].b1=d[i].b+d[i].k;
}
cin>>mm>>nn;
for(i=n;i>=1;i++)
{
if(mm>=d[i].a&&mm<=d[i].a1&&nn>=d[i].b&&nn<=d[i].b1)
{
cout<<i<<endl;
return 0;

}
if(i=1)
{
cout<<"-1"<<endl;
return 0;
}
}
}

0

yishengqi LV 6 @ 7 年前

#include <cstdio>
#include <iostream>
#include <cstdlib>
#define N 10000
using namespace std;
int main()
{
int n,a[N],b[N],x[N],y[N],h,l,t=-1;
cin>>n;
for(int i=0;i<n;i++)
{
cin>>a[i]>>b[i]>>x[i]>>y[i];
}
cin>>h>>l;
for(int i=0;i<n;i++)
{
if(h>=a[i]&&h<=(a[i]+x[i])&&l>=b[i]&&l<=(b[i]+y[i]))
t=i;
}
if(t==-1)
cout<<-1;
else
cout<<t+1;

return 0;
}

0

吴哥——传奇 LV 8 @ 7 年前

50 AC 纪念

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
using namespace std;
int main()
{
    int a[12223][10]={0},n,i,d=-1,x,y;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    {
        scanf("%d",&a[i][1]);
        scanf("%d",&a[i][2]);
        scanf("%d",&a[i][3]);
        scanf("%d\n",&a[i][4]);
    }
    scanf("%d %d",&x,&y);
    for(i=1;i<=n;i++)
    {
        if(x>=a[i][1]&&x<=a[i][1]+a[i][3]&&y>=a[i][2]&&y<=a[i][2]+a[i][4])
        d=i;
    }
    printf("%d",d);
}

你们的爸爸 @ 7 年前

不用开二维

0

Sky1231 (sky1231) LV 10 @ 7 年前

#include <cmath>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;

int n,x,y,ans;
int a[10000+1];
int b[10000+1];
int c[10000+1];
int d[10000+1];

int main()
{
    //freopen("carpet.in","r",stdin);
    //freopen("carpet.out","w",stdout);
    while (~scanf("%d",&n))
    {
        for (int i=1;i<=n;i++)
        {
            int d_x,d_y;
            scanf("%d%d%d%d",&a[i],&b[i],&d_x,&d_y);
            c[i]=a[i]+d_x,d[i]=b[i]+d_y;
        }
        ans=-1;
        scanf("%d%d",&x,&y);
        for (int i=1;i<=n;i++)
            if (a[i]<=x&&x<=c[i]&&b[i]<=y&&y<=d[i])
                ans=i;
        printf("%d\n",ans);
    }
}

sky1231 @ 7 年前

很H2O的题

0

bobby2002 LV 7 @ 7 年前

var
x,y,t,i,n:longint;
a,b,c,d:array[1..10000]of longint;
begin
readln(n);
for i:=1 to n do readln(a[i],b[i],c[i],d[i]);
readln(x,y);
for i:=n downto 1 do
if (x>=a[i])and(x<=a[i]+c[i])and(y>=b[i])and(y<=b[i]+d[i]) then begin
t:=i;
break;end;
if t=0 then write(-1) else write(t);
end.

0

bobby2002 LV 7 @ 7 年前

var
x,y,t,i,n:longint;
a,b,c,d:array[1..10000]of longint;
begin
readln(n);
for i:=1 to n do readln(a[i],b[i],c[i],d[i]);
readln(x,y);
for i:=n downto 1 do
if (x>=a[i])and(x<=a[i]+c[i])and(y>=b[i])and(y<=b[i]+d[i]) then begin
t:=i;
break;end;
if t=0 then write(-1) else write(t);
end.

0

沉江底 LV 9 @ 7 年前

唉，学了这么长时间也就只能AC点简单的模拟题了，55555

#include <iostream>
using namespace std;
int x[10003];
int y[10003];
int length[10003];
int width[10003];
int main(void)
{
    int n,flag=0,ans;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>x[i]>>y[i]>>length[i]>>width[i];
    }
    int p,q;
    cin>>p>>q;
    for(int i=0;i<n;i++)
    {
        if(x[i]+length[i]>=p&&y[i]+width[i]>=q&&x[i]<=p&&y[i]<=q)
        {
            flag=1;
            ans=i;
        }
    }
    if(flag)
    cout<<ans+1<<endl;
    else
    {
        cout<<"-1"<<endl;
    }
}

0

Commander_X LV 8 @ 8 年前

#include<cstdio>
#include<iostream>
#define MAX 100000

using namespace std;

int main(){
int n,a[MAX],b[MAX],g[MAX],k[MAX],x,y,i;
scanf("%d",&n);
for(i=0;i<n;i++){
scanf("%d%d%d%d",&a[i],&b[i],&g[i],&k[i]);
}
scanf("%d%d",&x,&y);
for(i=n-1;i>=-1;i--){
if(i==-1){
printf("-1");
break;
}
if((x>=a[i])&&(x<=a[i]+g[i])&&(y>=b[i])&&(y<=b[i]+k[i])){
printf("%d\n",i+1);
break;
}
}
return 0;
}

题解

60 条题解

状态耗时内存占用

铺地毯

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状态

开发

支持

题解

60 条题解

状态 耗时 内存占用

铺地毯

信息

状态

开发

支持

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状态耗时内存占用