Codeforces Beta Round #32 (Div. 2, Codeforces format)
http://codeforces.com/contest/32
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 }
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 }
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 }
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 }
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\n",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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