2019ICPC南京站题解

A题大水题。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define f(i,a,b) for(int i=a;i<=b;i++)
 4 #define scan(i) scanf("%d",&i)
 5 #define pf printf
 6 using namespace std;
 7
 8 int main()
 9 {
10     int n;
11     scan(n);
12     f(i,1,n){
13         int t;
14         scan(t);
15         if(t%2){
16             pf("%d\n",t/2+2);
17         }
18         else pf("%d\n",t/2+1);
19     }
20     return 0;
21 }

H题小水题，有两个wa点，第一是只有诚实的人且人数大于1人时直接输出1，第二是只有一个人且是诚实的时输出0。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define f(i,a,b) for(int i=a;i<=b;i++)
 4 #define scan(i) scanf("%d",&i)
 5 #define pf printf
 6 using namespace std;
 7
 8 int main()
 9 {
10     int a,b,c;
11     scanf("%d%d%d",&a,&b,&c);
12     if(b==0&&c==0){
13         if(a==1){
14             pf("YES\n0");
15         }
16         else pf("YES\n1");
17     }
18     else if(a>b+c){
19         pf("YES\n%d",2*b+2*c+1);
20     }
21     else pf("NO");
22     return 0;
23 }

K题计算几何。使用了一个假板子，发现了问题所在，已更新板子。

 1 #include <bits/stdc++.h>
 2 #define ll long long
 3 #define f(i,a,b) for(int i=a;i<=b;i++)
 4 #define scan(i) scanf("%d",&i)
 5 #define pf printf
 6 using namespace std;
 7 const  double eps=1e-10;
 8 const double PI=acos(-1.0);
 9 using namespace std;
10 struct Point{
11     double x;
12     double y;
13     Point(double x=0,double y=0):x(x),y(y){}
14     void operator<<(Point &A) {cout<<A.x<<‘ ‘<<A.y<<endl;}
15     bool operator!=(Point &A) {if(fabs(A.x-x)>eps||fabs(A.y-y)>eps) return true;return false;}
16 };
17
18 int dcmp(double x)  {return (x>eps)-(x<-eps); }
19 int sgn(double x)  {return (x>eps)-(x<-eps); }
20 typedef  Point  Vector;
21
22 Vector  operator +(Vector A,Vector B) { return Vector(A.x+B.x,A.y+B.y);}
23 Vector  operator -(Vector A,Vector B) { return Vector(A.x-B.x,A.y-B.y); }
24 Vector  operator *(Vector A,double p) { return Vector(A.x*p,A.y*p);  }
25 Vector  operator /(Vector A,double p) {return Vector(A.x/p,A.y/p);}
26 ostream &operator<<(ostream & out,Point & P) { out<<P.x<<‘ ‘<<P.y<<endl; return out;}
27 bool  operator< (const Point &A,const Point &B) { return dcmp(A.x-B.x)<0||(dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)<0); }
28 bool  operator== ( const Point &A,const Point &B) { return dcmp(A.x-B.x)==0&&dcmp(A.y-B.y)==0;}
29
30 double  Dot(Vector A,Vector B) {return A.x*B.x+A.y*B.y;}
31 double  Cross(Vector A,Vector B)  {return A.x*B.y-B.x*A.y; }
32 double  Length(Vector A)  { return sqrt(Dot(A, A));}
33 double  Angle(Vector A,Vector B) {return acos(Dot(A,B)/Length(A)/Length(B));}
34 double  Area2(Point A,Point B,Point C ) {return fabs(Cross(B-A, C-A));}
35
36 Vector Rotate(Vector A,double rad) { return Vector(A.x*cos(rad)-A.y*sin(rad),A.x*sin(rad)+A.y*cos(rad));}
37 Vector Normal(Vector A) {double L=Length(A);return Vector(-A.y/L,A.x/L);}
38
39 Point GetLineIntersection(Point P,Vector v,Point Q,Vector w){
40     if(v.x*w.y==v.y*w.x) return Point(5201314,5201314);
41     Vector u=P-Q;
42     double t=Cross(w, u)/Cross(v,w);
43     return P+v*t;
44 }//输入两个点斜式方程输出交点
45
46 bool  OnSegment(Point P,Point A,Point B){
47     return dcmp(Cross(P-A, P-B))==0&&dcmp(Dot(P-A,P-B))<=0;
48 }//输入三个点，判断P是否在线段AB上
49
50 double x1,x2,x3,x4,y11,y2,y3,y4;
51 int main()
52 {
53     int T;
54     scan(T);
55     f(kk,1,T){
56         scanf("%lf%lf%lf%lf%lf%lf%lf%lf",&x1,&y11,&x2,&y2,&x3,&y3,&x4,&y4);
57         Point a(x1,y11),b(x2,y2),c(x3,y3),p(x4,y4);
58         int x=0;
59         if(OnSegment(p,a,b)) x=1;
60         else if(OnSegment(p,a,c)) x=2;
61         else if(OnSegment(p,b,c)) x=3;
62         if(x==0) pf("-1\n");
63         else{
64             double area=Area2(a,b,c)/2;
65             if(x==2) swap(b,c);
66             else if(x==3) swap(a,c);
67             if(Length(p-a)>Length(p-b)){
68                 double angle=Angle(a-b,a-c);
69                 //area*0.5=0.5*a*b*sinx;
70                 double k=area/sin(angle)/Length(p-a)/Length(c-a);
71                 //cout<<k<<endl;
72                 Point ans=(c-a)*k+a;
73                 pf("%.10lf %.10lf\n",ans.x,ans.y);
74             }
75             else{
76                 double angle=Angle(b-a,b-c);
77                 //area*0.5=0.5*a*b*sinx;
78                 double k=area/sin(angle)/Length(p-b)/Length(c-b);
79                 //cout<<k<<endl;
80                 Point ans=(c-b)*k+b;
81                 pf("%.10lf %.10lf\n",ans.x,ans.y);
82             }
83         }
84     }
85 }

原文地址：https://www.cnblogs.com/St-Lovaer/p/11992417.html