2019-2020 ICPC Asia Taipei-Hsinchu Regional Contest 训练记录

Posted 2021-03-05 uuzhateteee

solved:9/13

upsolved:9/13

C.

签到题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int n;
 4 int a[55];
 5 int main()
 6 {
 7     scanf("%d",&n);
 8     for(int i=1;i<=n;++i)scanf("%d",&a[i]);
 9     bool yes=1;
10     for(int i=1;i<=n;++i)
11         for(int j=1;j<=n;++j)if(j!=i)
12             for(int k=1;k<=n;++k)if(k!=i&&k!=j)
13             {
14                 int x=abs(a[i]-a[j]);
15                 if(x%a[k])yes=0;    
16             }
17     if(yes)puts("yes");
18     else puts("no");
19 }

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D.

签到题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 string s;
 4 int main()
 5 {
 6     bool has=0;
 7     while(cin>>s)
 8     {
 9         if(s=="bubble"||s=="tapioka")continue;
10         has=1;
11         cout<<s<<" "; 
12     }
13     if(!has)puts("nothing");
14 }

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E.

构造一个长度1999的序列

考虑第一个数是负数，那么假算法一定不会包含这个数,取(a_1=-1)

序列为(-1,a_2,a_3,…,a_{1999})

那么假算法的解为(1998*sum_{i=2}^{1999}{a_i})

真算法的解为(1999*(sum_{i=2}^{1999}{a_i}-1))

那么假算法的解比真算法小$k$，则有(1999*(sum_{i=2}^{1999}{a_i}-1)=1998*sum_{i=2}^{1999}{a_i}+k)

则有(sum_{i=2}^{1999}{a_i}=1999+k)

随便构造一下凑够(k)就完事了

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int T,k,L;
 4 int main()
 5 {
 6     scanf("%d",&T);
 7     while(T--)
 8     {
 9         scanf("%d%d",&k,&L);
10         if(L>=2000)puts("-1");
11         else
12         {
13             puts("1999");
14             printf("-1 ");
15             int s=k+1999,x=1000000;
16             for(int i=1;i<=1998;++i)
17             {
18                 if(s>=x)
19                 {
20                     printf("%d ",x);
21                     s-=x;
22                 }
23                 else
24                 {
25                     printf("%d ",s);
26                     s=0;
27                 }
28             }
29         }
30     }
31 }

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H.

考虑方程(frac{1}{x}+frac{1}{y}=frac{1}{n})

移项通分有(frac{1}{y}=frac{x-n}{xn})

则(y=frac{xn}{x-n})

令(a=x-n)，则有(y=n+frac{n^2}{a})

那么枚举(n^2)的约数,解出(x,y)即可

然后(a=x oplus y)

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 int T;
 5 ll n;
 6 int main()
 7 {
 8     cin>>T;
 9     while(T--)
10     {
11         cin>>n;
12         ll ans=0;
13         for(ll i=1;i<=n;++i)if((n*n)%i==0)
14         {
15             ll y=n+n*n/i;
16             ll x=n+i;
17             ll a=x^y;
18             ans=max(ans,a);
19         }
20         cout<<ans<<endl;
21     }
22 }

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J.

二进制枚举集合，签到题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 int T,n,m;
 4 bitset<505> a[18];
 5 int main()
 6 {
 7     scanf("%d",&T);
 8     while(T--)
 9     {
10         scanf("%d%d",&n,&m);
11         for(int i=0;i<m;++i)
12         {
13             char str[505];
14             scanf("%s",str);
15             for(int j=0;j<n;++j)a[i][j]=0;
16             for(int j=0;j<n;++j)if(str[j]==‘1‘)a[i][j]=1;
17         }
18         int ans=m+1;
19         for(int S=0;S<(1<<m);++S)
20         {
21             bitset<505> b;
22             for(int i=0;i<n;++i)b[i]=0;
23             int cnt=0;
24             for(int i=0;i<m;++i)if(S&(1<<i))b|=a[i],cnt++;
25             bool yes=1;
26             for(int i=0;i<n;++i)if(!b[i])yes=0;
27             if(yes)ans=min(ans,cnt);
28         }
29         if(ans<=m)printf("%d
",ans);
30         else puts("-1");
31     }
32 }

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L.

因为可能是退化的四边形,所以不能在凸包上直接找点

考虑四边形的对角线一定是凸包上的对踵点，那么这步可以旋转卡壳得到对踵点

我们下面不能直接在凸包上三分了，但本题允许(O(n^2))的时间复杂度，那么我们枚举点

四边形可能是凹的，所以我们枚举的时候需要分别记录这条线两边最大最小的面积

如果某边没有点，就是最大减最小，否则就是两边相加

 1 #include<bits/stdc++.h>
 2 #define ll long long
 3 using namespace std;
 4 struct Point
 5 {
 6     ll x,y;
 7     Point(ll X=0,ll Y=0):x(X),y(Y){}
 8 };
 9 typedef Point Vector;
10 inline Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}
11 inline Vector operator - (Point A,Point B){return Vector(A.x-B.x, A.y-B.y);}
12 ll Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}
13 bool operator < (Point A,Point B){if(A.x==B.x)return A.y<B.y;return A.x<B.x;}
14 bool operator == (Point A,Point B){return (A.x==B.x)&&(A.y==B.y);}
15 vector<Point> ConvexHull(vector<Point>& p)
16 {
17     sort(p.begin(),p.end());
18     p.erase(unique(p.begin(),p.end()),p.end());
19     int n=p.size();
20     int m=0;
21     vector<Point> ch(n+1);
22     for(int i=0;i<n;++i)
23     {
24         while(m>1&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
25         ch[m++]=p[i];
26     }
27     int k=m;
28     for(int i=n-2;i>=0;--i)
29     {
30         while(m>k&&Cross(ch[m-1]-ch[m-2],p[i]-ch[m-2])<=0)m--;
31         ch[m++]=p[i];
32     }
33     if(n>1)m--;
34     ch.resize(m);
35     return ch;
36 }
37 int T,n,m;
38 vector<Point> A,H;
39 ll calc(Point U,Point V,bool tst=0)
40 {
41     bool firU=0,firV=0;
42     vector<Point> B;B.clear();
43     for(Point p:A)
44     {
45         if(p==U&&(!firU))firU=1;
46         else if(p==V&&(!firV))firV=1;
47         else B.push_back(p);
48     }
49     ll mx1=-1,mx2=-1,mn1=(ll)2e18,mn2=(ll)2e18;
50     for(Point p:B)
51     {
52         //if(tst)printf("(%I64d,%I64d)
",p.x,p.y);
53         if(Cross(V-U,p-U)<0)
54         {
55             mx1=max(mx1,abs(Cross(V-U,p-U)));
56             mn1=min(mn1,abs(Cross(V-U,p-U)));
57         }
58         else
59         {
60             mx2=max(mx2,abs(Cross(V-U,p-U)));
61             mn2=min(mn2,abs(Cross(V-U,p-U)));
62         }
63     }
64     if(mx1==-1)return mx2-mn2;
65     else if(mx2==-1)return mx1-mn1;
66     else return mx1+mx2;
67 }
68 int main()
69 {
70     scanf("%d",&T);
71     while(T--)
72     {
73         A.clear();
74         scanf("%d",&n);
75         for(int i=1;i<=n;++i)
76         {
77             ll x,y;
78             scanf("%I64d%I64d",&x,&y);
79             A.push_back(Point(x,y));
80         }
81         H=A;
82         H=ConvexHull(H);
83         m=H.size();
84         if(m<=2)
85         {
86             puts("0");
87             continue;
88         }
89         int j=1;
90         ll ans=0;
91         for(int i=0;i<m;++i)
92         {
93             while(abs(Cross(H[j]-H[i],H[j]-H[(i+1)%m]))<abs(Cross(H[(j+1)%m]-H[i],H[(j+1)%m]-H[(i+1)%m])))j=(j+1)%m;
94             ans=max(ans,calc(H[i],H[j]));
95         }
96         if(ans&1)printf("%I64d.5
",ans/2);
97         else printf("%I64d
",ans/2);
98     }
99 }

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