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\n",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)\n",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\n",ans/2);
97 else printf("%I64d\n",ans/2);
98 }
99 }
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