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Codeforces Beta Round #32 (Div. 2, Codeforces format)

A

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 /*#ifndef ONLINE_JUDGE
11         freopen("1.txt","r",stdin);
12 #endif */
13 
14 int n,m;
15 int a[1005];
16 
17 int main(){
18     #ifndef ONLINE_JUDGE
19       //  freopen("1.txt","r",stdin);
20     #endif
21     std::ios::sync_with_stdio(false);
22     cin>>n>>m;
23     int ans=0;
24     for(int i=1;i<=n;i++) cin>>a[i];
25     for(int i=1;i<=n;i++){
26         for(int j=1;j<=n;j++){
27             if(i!=j&&abs(a[i]-a[j])<=m){
28                 ans++;
29             }
30         }
31     }
32     cout<<ans<<endl;
33 }

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B

模拟

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 /*#ifndef ONLINE_JUDGE
11         freopen("1.txt","r",stdin);
12 #endif */
13 
14 int n,m;
15 int a[1005];
16 
17 int main(){
18     #ifndef ONLINE_JUDGE
19       //  freopen("1.txt","r",stdin);
20     #endif
21     std::ios::sync_with_stdio(false);
22     string str;
23     cin>>str;
24     for(int i=0;i<str.length();i++){
25         if(str[i]=='.'){
26             cout<<0;
27         }
28         else{
29             if(str[i+1]=='.'){
30                 cout<<1;
31             }
32             else{
33                 cout<<2;
34             }
35             i++;
36         }
37     }
38 }

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C

水题

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 /*#ifndef ONLINE_JUDGE
11         freopen("1.txt","r",stdin);
12 #endif */
13 
14 
15 int main(){
16     #ifndef ONLINE_JUDGE
17       //  freopen("1.txt","r",stdin);
18     #endif
19     std::ios::sync_with_stdio(false);
20     ll n,m,s;
21     cin >> n  >> m >> s;
22     ll w = (n - 1) / s;
23     ll h = (m - 1) / s;
24     ll cnt = (w + 1) * (h + 1);
25     ll a = min (s - 1 , (m - 1) % s);
26     ll b = min (s - 1 , (n - 1) % s);
27     cout << (a + 1) * (b + 1) * cnt << endl;
28     return 0;
29 }

View Code

D

模拟

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define lson l,mid,rt<<1
 4 #define rson mid+1,r,rt<<1|1
 5 #define sqr(x) ((x)*(x))
 6 #define pb push_back
 7 #define maxn 1000005
 8 typedef long long ll;
 9 typedef unsigned long long ull;
10 /*#ifndef ONLINE_JUDGE
11         freopen("1.txt","r",stdin);
12 #endif */
13 
14 string a[305];
15 int n,m,k;
16 int main(){
17     #ifndef ONLINE_JUDGE
18       //  freopen("1.txt","r",stdin);
19     #endif
20     std::ios::sync_with_stdio(false);
21     cin>>n>>m>>k;
22     for (int i=0;i<n;i++) cin>>a[i];
23     for (int r=1;r<n;r++)
24         for (int i=r;i<n-r;i++)
25             for (int j=r;j<m-r;j++)
26                 if (a[i][j]=='*'&&a[i+r][j]=='*'&&a[i][j+r]=='*'&&a[i-r][j]=='*'&&a[i][j-r]=='*'){
27                     k--;
28                     if (!k){
29                         cout<<i+1<<" "<<j+1<<" "<<i-r+1<<" "<<j+1<<" "<<i+r+1<<" "<<j+1<<" "<<i+1<<" "<<j-r+1<<" "<<i+1<<" "<<j+r+1<<endl;
30                         return 0;
31                     }
32                 }
33     cout<<-1<<endl;
34 }

View Code

E

几何

先判断AB是否可以直接相连，不能的话就做A对于镜面的对称点，连接对称点和B,判断这条线段和镜面是否有交点，再判断交点连A是否会经过墙，交点连B是否会经过墙

  1 #include<bits/stdc++.h>
  2 using namespace std;
  3 #define lson l,mid,rt<<1
  4 #define rson mid+1,r,rt<<1|1
  5 #define sqr(x) ((x)*(x))
  6 #define pb push_back
  7 #define eps 1e-8
  8 #define PI acos(-1.0)
  9 #define maxn 1000005
 10 typedef long long ll;
 11 typedef unsigned long long ull;
 12 
 13 /*#ifndef ONLINE_JUDGE
 14         freopen("1.txt","r",stdin);
 15 #endif */
 16 int sgn(double x){
 17     if(fabs(x)<eps) return 0;
 18     if(x<0) return -1;
 19     else return 1;
 20 }
 21 struct Point{
 22     double x,y;
 23     Point(){}
 24     Point(double _x,double _y){
 25         x=_x;
 26         y=_y;
 27     }
 28     void input(){
 29         scanf("%lf %lf",&x,&y);
 30     }
 31     void output(){
 32         printf("%.2f %.2f
",x,y);
 33     }
 34     bool operator == (const Point &b)const{
 35         return sgn(x-b.x) == 0 && sgn(y-b.y)== 0;
 36     }
 37     bool operator < (const Point &b)const{
 38         return sgn(x-b.x)==0?sgn(y-b.y)<0:x<b.x;
 39     }
 40     Point operator - (const Point &b)const{
 41         return Point(x-b.x,y-b.y);
 42     }
 43     //叉积
 44     double operator ^ (const Point &b)const{
 45         return x*b.y-y*b.x;
 46     }
 47     //点积
 48     double operator * (const Point &b)const{
 49         return x*b.x+y*b.y;
 50     }
 51     //返回长度
 52     double len(){
 53         return hypot(x,y);
 54     }
 55     //返回长度的平方
 56     double len2(){
 57         return x*x+y*y;
 58     }
 59     //返回两点的距离
 60     double distance(Point p){
 61         return hypot(x-p.x,y-p.y);
 62     }
 63     Point operator + (const Point &b)const{
 64         return Point(x+b.x,y+b.y);
 65     }
 66     Point operator * (const double &k)const{
 67         return Point(x*k,y*k);
 68     }
 69     Point operator / (const double &k)const{
 70         return Point(x/k,y/k);
 71     }
 72 
 73     //计算pa和pb的夹角
 74     //就是求这个点看a,b所成的夹角
 75     ///LightOJ1202
 76     double rad(Point a,Point b){
 77         Point p=*this;
 78         return fabs(atan2(fabs((a-p)^(b-p)),(a-p)*(b-p)));
 79     }
 80     //化为长度为r的向量
 81     Point trunc(double r){
 82         double l=len();
 83         if(!sgn(l)) return *this;
 84         r/=l;
 85         return Point(x*r,y*r);
 86     }
 87     //逆时针转90度
 88     Point rotleft(){
 89         return Point(-y,x);
 90     }
 91     //顺时针转90度
 92     Point rotright(){
 93         return Point(y,-x);
 94     }
 95     //绕着p点逆时针旋转angle
 96     Point rotate(Point p,double angle){
 97         Point v=(*this) -p;
 98         double c=cos(angle),s=sin(angle);
 99         return Point(p.x+v.x*c-v.y*s,p.y+v.x*s+v.y*c);
100     }
101 };
102 
103 struct Line{
104     Point s,e;
105     Line(){}
106     Line(Point _s,Point _e){
107         s=_s;
108         e=_e;
109     }
110     bool operator==(Line v){
111         return (s==v.s)&&(e==v.e);
112     }
113     //根据一个点和倾斜角angle确定直线，0<=angle<pi
114     Line(Point p,double angle){
115         s=p;
116         if(sgn(angle-PI/2)==0){
117             e=(s+Point(0,1));
118         }
119         else{
120             e=(s+Point(1,tan(angle)));
121         }
122     }
123     //ax+by+c=0;
124     Line(double a,double b,double c){
125         if(sgn(a)==0){
126             s=Point(0,-c/b);
127             e=Point(1,-c/b);
128         }
129         else if(sgn(b)==0){
130             s=Point(-c/a,0);
131             e=Point(-c/a,1);
132         }
133         else{
134             s=Point(0,-c/b);
135             e=Point(1,(-c-a)/b);
136         }
137     }
138     void input(){
139         s.input();
140         e.input();
141     }
142     void adjust(){
143         if(e<s) swap(s,e);
144     }
145     //求线段长度
146     double length(){
147         return s.distance(e);
148     }
149     inline double operator * (const Point &p) const {
150         return (e - s) * (p - s);
151     }
152     //返回直线倾斜角 0<=angle<pi
153     double angle(){
154         double k=atan2(e.y-s.y,e.x-s.x);
155         if(sgn(k)<0) k+=PI;
156         if(sgn(k-PI)==0) k-=PI;
157         return k;
158     }
159     //点和直线的关系
160     //1 在左侧
161     //2 在右侧
162     //3 在直线上
163     int relation(Point p){
164         int c=sgn((p-s)^(e-s));
165         if(c<0) return 1;
166         else if(c>0) return 2;
167         else return 3;
168     }
169     //点在线段上的判断
170     bool pointonseg(Point p){
171         return sgn((p-s)^(e-s))==0&&sgn((p-s)*(p-e))<=0;
172     }
173     //两向量平行(对应直线平行或重合)
174     bool parallel(Line v){
175       //  cout<<sgn((e-s)^(v.e-v.s))<<endl;
176         return sgn((e-s)^(v.e-v.s))==0;
177     }
178     //两线段相交判断
179     //2 规范相交
180     //1 非规范相交
181     //0 不相交
182     int segcrossseg(Line v){
183         int d1=sgn((e-s)^(v.s-s));
184         int d2=sgn((e-s)^(v.e-s));
185         int d3=sgn((v.e-v.s)^(s-v.s));
186         int d4=sgn((v.e-v.s)^(e-v.s));
187         if((d1^d2)==-2&&(d3^d4)==-2) return 2;
188         return (d1==0&&sgn((v.s-s)*(v.s-e))<=0||
189                 d2==0&&sgn((v.e-s)*(v.e-e))<=0||
190                 d3==0&&sgn((s-v.s)*(s-v.e))<=0||
191                 d4==0&&sgn((e-v.s)*(e-v.e))<=0);
192     }
193     //直线和线段相交判断
194     //-*this line -v seg
195     //2 规范相交
196     //1 非规范相交
197     //0 不相交
198     int linecrossseg(Line v){
199         int d1=sgn((e-s)^(v.s-s));
200         int d2=sgn((e-s)^(v.e-s));
201         if((d1^d2)==-2) return 2;
202         return (d1==0||d2==0);
203     }
204     //两直线关系
205     //0 平行
206     //1 重合
207     //2 相交
208     int linecrossline(Line v){
209         if((*this).parallel(v))
210             return v.relation(s)==3;
211         return 2;
212     }
213     //求两直线的交点
214     //要保证两直线不平行或重合
215     Point crosspoint(Line v){
216         double a1=(v.e-v.s)^(s-v.s);
217         double a2=(v.e-v.s)^(e-v.s);
218         if(a1==a2) return Point(1e9,1e9);
219         return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
220     }
221     Point PointtoLine(const Point &p){
222         return crosspoint(Line(p,p+(s-e).rotleft()));
223     }
224     Point SymPoint (const Point &p) {
225         return PointtoLine (p) * 2 - p;
226     }
227     //点到直线的距离
228     double dispointtoline(Point p){
229         return fabs((p-s)^(e-s))/length();
230     }
231     //点到线段的距离
232     double dispointtoseg(Point p){
233         if(sgn((p-s)*(e-s))<0||sgn((p-e)*(s-e))<0)
234             return min(p.distance(s),p.distance(e));
235         return dispointtoline(p);
236     }
237     //返回线段到线段的距离
238     //前提是两线段不相交，相交距离就是0了
239     double dissegtoseg(Line v){
240         return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
241     }
242     //返回点P在直线上的投影
243     Point lineprog(Point p){
244         return s+(((e-s)*((e-s)*(p-s)))/((e-s).len2()));
245     }
246     //返回点P关于直线的对称点
247     Point symmetrypoint(Point p){
248         Point q=lineprog(p);
249         return Point(2*q.x-p.x,2*q.y-p.y);
250     }
251 };
252 ///两点之间，两点和镜子之间，镜子和墙
253 int main(){
254     #ifndef ONLINE_JUDGE
255         freopen("1.txt","r",stdin);
256     #endif
257     std::ios::sync_with_stdio(false);
258     Point v,p;
259     v.input();
260     p.input();
261     Line w,m,L;
262     w.input();
263     m.input();
264     L.s=v,L.e=p;
265     if(L.segcrossseg(w)==0){
266         if(L.segcrossseg(m)==0||L.parallel(m)){
267             cout<<"YES"<<endl;
268             return 0;
269         }
270     }
271     else{
272         Point tt=m.symmetrypoint(p);
273         if(m.segcrossseg(Line(v,tt))!=0){
274             Point pp=m.crosspoint(Line(v,tt));
275             if(pp.x==1e9&&pp.y==1e9){
276                 pp=p;
277             }
278             Line l1,l2;
279             l1.s=v,l1.e=pp;
280             l2.s=p,l2.e=pp;
281            // cout<<l1.segcrossseg(w)<<" "<<l2.segcrossseg(w)<<endl;
282             if(l1.segcrossseg(w)==0&&l2.segcrossseg(w)==0){
283                 cout<<"YES"<<endl;
284                 return 0;
285             }
286         }
287     }
288     cout<<"NO"<<endl;
289     return 0;
290 }

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原文地址：https://www.cnblogs.com/Fighting-sh/p/10389581.html