2019ICPC南京站题解
作者：互联网
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 }
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