Angle Beats Gym - 102361A
作者:互联网
题目链接:https://vjudge.net/problem/Gym-102361A
题意:给定N个点,q次询问每次询问给一个点,问在N个点中取2个和给定点最多可以组成几个直角三角形。
思路:https://www.cnblogs.com/Jiaaaaaaaqi/p/11631203.html
1 #include <bits/stdc++.h> 2 #include <time.h> 3 #include <set> 4 #include <map> 5 #include <stack> 6 #include <cmath> 7 #include <queue> 8 #include <cstdio> 9 #include <string> 10 #include <vector> 11 #include <cstring> 12 #include <utility> 13 #include <cstring> 14 #include <iostream> 15 #include <algorithm> 16 #include <list> 17 using namespace std; 18 //cout<<setprecision(10)<<fixed; 19 #define eps 1e-6 20 #define PI acos(-1.0) 21 #define lowbit(x) ((x)&(-x)) 22 #define zero(x) (((x)>0?(x):-(x))<eps) 23 #define mem(s,n) memset(s,n,sizeof s); 24 #define rep(i,a,b) for(int i=a;i<=b;i++) 25 #define rep2(i,a,b) for(int i=a;i>=b;i--) 26 #define ios {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);} 27 typedef long long ll; 28 typedef unsigned long long ull; 29 const int maxn=1e6+5; 30 const ll Inf=0x7f7f7f7f7f7f7f7f; 31 const ll mod=1e6+3; 32 //const int N=3e3+5; 33 bool isPowerOfTwo(int n) { return n > 0 && (n & (n - 1)) == 0; }//判断一个数是不是 2 的正整数次幂 34 int modPowerOfTwo(int x, int mod) { return x & (mod - 1); }//对 2 的非负整数次幂取模 35 int getBit(int a, int b) { return (a >> b) & 1; }// 获取 a 的第 b 位,最低位编号为 0 36 int Max(int a, int b) { return b & ((a - b) >> 31) | a & (~(a - b) >> 31); }// 如果 a>=b,(a-b)>>31 为 0,否则为 -1 37 int Min(int a, int b) { return a & ((a - b) >> 31) | b & (~(a - b) >> 31); } 38 ll gcd(ll a, ll b) {return b ? gcd(b, a % b) : a;} 39 ll lcm(ll a, ll b) {return a / gcd(a, b) * b;} 40 inline int read() 41 { 42 int X=0; bool flag=1; char ch=getchar(); 43 while(ch<'0'||ch>'9') {if(ch=='-') flag=0; ch=getchar();} 44 while(ch>='0'&&ch<='9') {X=(X<<1)+(X<<3)+ch-'0'; ch=getchar();} 45 if(flag) return X; 46 return ~(X-1); 47 } 48 inline void write(int X) 49 { 50 if(X<0) {X=~(X-1); putchar('-');} 51 if(X>9) write(X/10); 52 putchar(X%10+'0'); 53 } 54 /* 55 inline int write(int X) 56 { 57 if(X<0) {putchar('-'); X=~(X-1);} 58 int s[20],top=0; 59 while(X) {s[++top]=X%10; X/=10;} 60 if(!top) s[++top]=0; 61 while(top) putchar(s[top--]+'0'); 62 } 63 */ 64 int Abs(int n) { 65 return (n ^ (n >> 31)) - (n >> 31); 66 /* n>>31 取得 n 的符号,若 n 为正数,n>>31 等于 0,若 n 为负数,n>>31 等于 -1 67 若 n 为正数 n^0=n, 数不变,若 n 为负数有 n^(-1) 68 需要计算 n 和 -1 的补码,然后进行异或运算, 69 结果 n 变号并且为 n 的绝对值减 1,再减去 -1 就是绝对值 */ 70 } 71 ll binpow(ll a, ll b) { 72 ll res = 1; 73 while (b > 0) { 74 if (b & 1) res = res * a%mod; 75 a = a * a%mod; 76 b >>= 1; 77 } 78 return res%mod; 79 } 80 void extend_gcd(ll a,ll b,ll &x,ll &y) 81 { 82 if(b==0) { 83 x=1,y=0; 84 return; 85 } 86 extend_gcd(b,a%b,x,y); 87 ll tmp=x; 88 x=y; 89 y=tmp-(a/b)*y; 90 } 91 ll mod_inverse(ll a,ll m) 92 { 93 ll x,y; 94 extend_gcd(a,m,x,y); 95 return (m+x%m)%m; 96 } 97 ll eulor(ll x) 98 { 99 ll cnt=x; 100 ll ma=sqrt(x); 101 for(int i=2;i<=ma;i++) 102 { 103 if(x%i==0) cnt=cnt/i*(i-1); 104 while(x%i==0) x/=i; 105 } 106 if(x>1) cnt=cnt/x*(x-1); 107 return cnt; 108 } 109 struct node 110 { 111 ll x,y; 112 int id; 113 }p[maxn],be[maxn]; 114 int n,m; 115 int ans[maxn]; 116 int cmp1(node a,node b) 117 { 118 ll d=a.x*b.y-b.x*a.y; 119 if(d==0) return a.x<b.x; 120 else return d>0; 121 } 122 int qua(node a) 123 { 124 if(a.x>0&&a.y>=0) return 1; 125 else if(a.x<=0&&a.y>0) return 2; 126 else if(a.x<0&&a.y<=0) return 3; 127 else if(a.x>=0&&a.y<=0) return 4; 128 } 129 int cmp(node a,node b) 130 { 131 if(qua(a)==qua(b)) return cmp1(a,b); 132 else return qua(a)<qua(b); 133 } 134 ll check(node a,node b) 135 { 136 return a.x*b.x+a.y*b.y; 137 } 138 ll chaji(node a,node b) 139 { 140 return a.x*b.y-b.x*a.y; 141 } 142 ll work(node pp) 143 { 144 for(int i=1;i<=n;i++) 145 { 146 p[i]=be[i]; 147 p[i].x-=pp.x; 148 p[i].y-=pp.y; 149 } 150 p[0]=pp; 151 sort(p+1,p+n+1,cmp); 152 for(int i=1;i<=n;i++) 153 { 154 p[i+n]=p[i]; 155 } 156 ll ans=0; 157 int r=2; 158 for(int l=1;l<=n;l++) 159 { 160 while(r<=2*n) 161 { 162 if(chaji(p[l],p[r])<0) break; 163 if(check(p[l],p[r])<=0) break; 164 r++; 165 } 166 int tr=r; 167 while(tr<=2*n) 168 { 169 if(chaji(p[l],p[tr])<=0) break; 170 if(check(p[l],p[tr])!=0) break; 171 ans++; 172 tr++; 173 } 174 } 175 return ans; 176 } 177 int main() 178 { 179 while(~scanf("%d%d",&n,&m)) 180 { 181 int all=0; 182 for(int i=1;i<=n;i++) 183 { 184 all++; 185 int x,y; 186 scanf("%d%d",&x,&y); 187 p[all].x=x,p[all].y=y,p[all].id=0; 188 be[all]=p[all]; 189 } 190 for(int i=1;i<=m;i++) 191 { 192 all++; 193 int x,y; 194 scanf("%d%d",&x,&y); 195 p[all].x=x,p[all].y=y,p[all].id=i; 196 be[all]=p[all]; 197 ans[i]=work(p[all]); 198 } 199 for(int i=1;i<=n;i++) 200 { 201 for(int j=1;j<=all;j++) 202 { 203 p[j]=be[j]; 204 } 205 p[0]=be[i]; 206 int flag=0; 207 for(int j=1;j<=all;j++) 208 { 209 if(p[j].x==p[0].x&&p[j].y==p[0].y) flag=1; 210 if(flag) p[j]=p[j+1]; 211 p[j].x-=p[0].x; 212 p[j].y-=p[0].y; 213 } 214 int nn=all-1; 215 sort(p+1,p+1+nn,cmp); 216 for(int j=1;j<=nn;j++) 217 { 218 p[j+nn]=p[j]; 219 } 220 int r=2; 221 for(int l=1;l<=nn;l++) 222 { 223 int id=0; 224 if(p[0].id) id=p[0].id; 225 if(p[l].id) id=p[l].id; 226 while(r<=2*nn) 227 { 228 if(chaji(p[l],p[r])<0) break; 229 if(check(p[l],p[r])<=0) break; 230 r++; 231 } 232 int tr=r; 233 while(tr<=2*nn) 234 { 235 if(chaji(p[l],p[tr])<=0) break; 236 if(check(p[l],p[tr])!=0) break; 237 if(id==0) 238 { 239 if(p[tr].id) ans[p[tr].id]++; 240 } 241 else 242 { 243 if(p[tr].id==0) ans[id]++; 244 } 245 tr++; 246 } 247 } 248 } 249 for(int i=1;i<=m;i++) 250 { 251 printf("%d\n",ans[i]); 252 } 253 } 254 return 0; 255 }
标签:Angle,int,ll,Beats,31,return,include,Gym,mod 来源: https://www.cnblogs.com/zpj61/p/13625227.html