题解

60 条题解

  • 5
    @ 2017-03-10 20:09:37

    #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;
    }

    • @ 2017-08-25 01:45:55

      太复杂了,还是我的好

    • @ 2018-07-25 15:09:49

      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
    @ 2017-11-01 15:50:04

    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
    @ 2021-09-04 14:04:31
    #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
    @ 2018-12-16 19:07:45

    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
    @ 2018-02-06 10:05:39
    #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
    @ 2022-08-30 09:58:17
    #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
    @ 2022-08-30 09:57:55

    #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
    @ 2020-04-06 14:39:23
    #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
    @ 2018-07-25 15:08:01

    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
    @ 2018-05-16 16:33:30
    #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
    @ 2017-10-15 14:24:17

    ......

    #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
    @ 2017-10-10 23:30:00

    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
    @ 2017-10-03 11:47:41

    #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
    @ 2017-09-26 15:56:26

    #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
    @ 2017-08-25 01:45:06

    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);
    }
    
  • 0
    @ 2017-08-14 11:43:01
    #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);
        }
    }
    
  • 0
    @ 2017-07-12 21:37:56

    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
    @ 2017-07-12 21:37:28

    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
    @ 2017-06-16 21:34:28

    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
    @ 2017-04-26 18:38:25

    #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;
    }

信息

ID
1736
难度
4
分类
搜索 | 枚举 点击显示
标签
递交数
4963
已通过
2200
通过率
44%
被复制
13
上传者