2019杭电多校第五场
比赛记录
@byf水题写错一发，边界没考虑，变量还写错，一发罚时加很晚才过。1004忘记结构体重载运算符中的gcd也带一个log，TLE2发，而且这题花了太多时间。
@shl exkmp板子想不起来，先用后缀数组T了一发
@lfw 一上场用kmp写exkmp的题，结果发现不太对，有点冲动了，应该能推出样例再写的。1005一开始没想清楚，计数有问题，所以花了很久时间用对样例的调试来想清楚怎么计数。
题解
1001 fraction
题解：https://blog.csdn.net/baiyifeifei/article/details/98606715
#include<bits/stdc++.h> using namespace std; typedef long long LL; LL p,a; //solve "pa/pb<x/y<=qa/qb" 's min(y),min(x) void solve(LL pa,LL pb,LL qa,LL qb,LL &x,LL &y) { LL z=(pa+pb-1)/pb; if(z<=qa/qb){ x=z;y=1; return ; } pa-=(z-1)*pb;qa-=(z-1)*qb; solve(qb,qa,pb,pa,y,x); x+=(z-1)*y; } int main() { int t; scanf("%d",&t); LL x,y; while(t--) { scanf("%lld%lld",&p,&a); solve(p,a,p,a-1,x,y); printf("%lld/%lld\n",x*a-p*y,x); } }
1002 three arrays
题解：https://blog.csdn.net/liufengwei1/article/details/98536161
1 #include<bits/stdc++.h> 2 #define maxl 100010 3 using namespace std; 4 5 int n,m,tota,totb,sum; 6 int ans[maxl],mi[30]; 7 int a[maxl],b[maxl]; 8 int suma[maxl*31],sumb[maxl*31]; 9 int tra[maxl*31][2],trb[maxl*31][2]; 10 int q[maxl],w[maxl]; 11 vector <int> numa[maxl*31],numb[maxl*31]; 12 vector <int> :: iterator it; 13 bool ina[maxl],inb[maxl]; 14 15 inline void inserta(int id,int x) 16 { 17 int u=0,c; 18 for(int i=29;i>=0;i--) 19 { 20 suma[u]++; 21 c=(x>>i)&1; 22 if(!tra[u][c]) 23 tra[u][c]=++tota; 24 u=tra[u][c]; 25 } 26 suma[u]++;numa[u].push_back(id); 27 } 28 29 inline void insertb(int id,int x) 30 { 31 int u=0,c; 32 for(int i=29;i>=0;i--) 33 { 34 sumb[u]++; 35 c=(x>>i)&1; 36 if(!trb[u][c]) 37 trb[u][c]=++totb; 38 u=trb[u][c]; 39 } 40 sumb[u]++;numb[u].push_back(id); 41 } 42 43 inline void prework() 44 { 45 for(int i=0;i<=tota;i++) 46 { 47 tra[i][0]=tra[i][1]=0; 48 suma[i]=0;numa[i].clear(); 49 } 50 for(int i=0;i<=totb;i++) 51 { 52 trb[i][0]=trb[i][1]=0; 53 sumb[i]=0,numb[i].clear(); 54 } 55 scanf("%d",&n); 56 tota=0;totb=0; 57 for(int i=1;i<=n;i++) 58 { 59 scanf("%d",&a[i]); 60 inserta(i,a[i]);ina[i]=true; 61 } 62 for(int i=1;i<=n;i++) 63 { 64 scanf("%d",&b[i]); 65 insertb(i,b[i]);inb[i]=true; 66 } 67 } 68 69 inline int finda(int x) 70 { 71 int u=0,c,tmp=0; 72 for(int i=29;i>=0;i--) 73 { 74 c=(x>>i)&1; 75 if(!tra[u][c]) 76 u=tra[u][c^1]; 77 else 78 u=tra[u][c]; 79 } 80 it=numa[u].end();--it; 81 return (*it); 82 } 83 84 inline int findb(int x) 85 { 86 int u=0,c; 87 for(int i=29;i>=0;i--) 88 { 89 c=(x>>i)&1; 90 if(!trb[u][c]) 91 u=trb[u][c^1]; 92 else 93 u=trb[u][c]; 94 } 95 it=numb[u].end();--it; 96 return (*it); 97 } 98 99 inline void dela(int x) 100 { 101 int u=0,c,last;suma[u]--; 102 for(int i=29;i>=0;i--) 103 { 104 c=(x>>i)&1;last=u; 105 u=tra[u][c]; 106 suma[u]--; 107 if(suma[u]==0) 108 tra[last][c]=0; 109 } 110 it=numa[u].end();--it; 111 numa[u].erase(it); 112 } 113 114 inline void delb(int x) 115 { 116 int u=0,c,last;sumb[u]--; 117 for(int i=29;i>=0;i--) 118 { 119 c=(x>>i)&1;last=u; 120 u=trb[u][c]; 121 sumb[u]--; 122 if(sumb[u]==0) 123 trb[last][c]=0; 124 } 125 it=numb[u].end();--it; 126 numb[u].erase(it); 127 } 128 129 inline void dfs(int k) 130 { 131 int id; 132 if(k&1) 133 { 134 id=finda(b[q[k-1]]); 135 if(id==q[k-2] && k-2>0) 136 { 137 delb(b[q[k-1]]); 138 dela(a[id]); 139 inb[q[k-1]]=false; 140 ina[id]=false; 141 ans[++ans[0]]=b[q[k-1]]^a[id]; 142 k-=2;sum--; 143 } 144 else 145 { 146 w[k]=b[q[k-1]]^a[id]; 147 q[k]=id;k++; 148 } 149 if(sum>0) 150 dfs(k); 151 } 152 else 153 { 154 id=findb(a[q[k-1]]); 155 if(id==q[k-2] && k-2>0) 156 { 157 dela(a[q[k-1]]); 158 delb(b[id]); 159 ina[q[k-1]]=false; 160 inb[id]=false; 161 ans[++ans[0]]=a[q[k-1]]^b[id]; 162 k-=2;sum--; 163 } 164 else 165 { 166 w[k]=a[q[k-1]]^b[id]; 167 q[k]=id;k++; 168 } 169 if(sum>0) 170 dfs(k); 171 } 172 } 173 174 inline void mainwork() 175 { 176 ans[0]=0;sum=n; 177 for(int i=1;i<=n;i++) 178 if(ina[i]) 179 { 180 q[1]=i; 181 dfs(2); 182 } 183 } 184 185 inline void print() 186 { 187 sort(ans+1,ans+1+ans[0]); 188 for(int i=1;i<=n;i++) 189 printf("%d%c",ans[i],(i==n)?'\n':' '); 190 } 191 192 int main() 193 { 194 mi[0]=1; 195 for(int i=1;i<=29;i++) 196 mi[i]=mi[i-1]*2; 197 int t; 198 scanf("%d",&t); 199 for(int i=1;i<=t;i++) 200 { 201 prework(); 202 mainwork(); 203 print(); 204 } 205 return 0; 206 }
1003 geometric problem
unsolved
1004 equation
1 #include<bits/stdc++.h> 2 using namespace std; 3 4 namespace IO{ 5 #define BUF_SIZE 100000 6 #define OUT_SIZE 100000 7 #define ll long long 8 9 bool IOerror=0; 10 inline char nc(){ 11 static char buf[BUF_SIZE],*p1=buf+BUF_SIZE,*pend=buf+BUF_SIZE; 12 if (p1==pend){ 13 p1=buf; pend=buf+fread(buf,1,BUF_SIZE,stdin); 14 if (pend==p1){IOerror=1;return -1;} 15 } 16 return *p1++; 17 } 18 inline bool blank(char ch){return ch==' '||ch=='\n'||ch=='\r'||ch=='\t';} 19 inline bool read(int &x){ 20 bool sign=0; char ch=nc(); x=0; 21 for (;blank(ch);ch=nc()); 22 if (IOerror) return false; 23 if (ch=='-')sign=1,ch=nc(); 24 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 25 if (sign)x=-x; return true; 26 } 27 inline bool read(ll &x){ 28 bool sign=0; char ch=nc(); x=0; 29 for (;blank(ch);ch=nc()); 30 if (IOerror) return false; 31 if (ch=='-')sign=1,ch=nc(); 32 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 33 if (sign)x=-x; return true; 34 } 35 inline bool read(double &x){ 36 bool sign=0; char ch=nc(); x=0; 37 for (;blank(ch);ch=nc()); 38 if (IOerror) return false; 39 if (ch=='-')sign=1,ch=nc(); 40 for (;ch>='0'&&ch<='9';ch=nc())x=x*10+ch-'0'; 41 if (ch=='.'){ 42 double tmp=1; ch=nc(); 43 for (;ch>='0'&&ch<='9';ch=nc())tmp/=10.0,x+=tmp*(ch-'0'); 44 } 45 if (sign)x=-x; return true; 46 } 47 inline bool read(char *s){ 48 char ch=nc(); 49 for (;blank(ch);ch=nc()); 50 if (IOerror) return false; 51 for (;!blank(ch)&&!IOerror;ch=nc())*s++=ch; 52 *s=0; return true; 53 } 54 inline void read(char &c){ 55 for (c=nc();blank(c);c=nc()); 56 if (IOerror){c=-1;return;} 57 } 58 //fwrite->write 59 struct Ostream_fwrite{ 60 char *buf,*p1,*pend; 61 Ostream_fwrite(){buf=new char[BUF_SIZE];p1=buf;pend=buf+BUF_SIZE;} 62 void out(char ch){ 63 if (p1==pend){ 64 fwrite(buf,1,BUF_SIZE,stdout);p1=buf; 65 } 66 *p1++=ch; 67 } 68 void print(int x){ 69 static char s[15],*s1;s1=s; 70 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 71 while(x)*s1++=x%10+'0',x/=10; 72 while(s1--!=s)out(*s1); 73 } 74 void println(int x){ 75 static char s[15],*s1;s1=s; 76 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 77 while(x)*s1++=x%10+'0',x/=10; 78 while(s1--!=s)out(*s1); out('\n'); 79 } 80 void print(ll x){ 81 static char s[25],*s1;s1=s; 82 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 83 while(x)*s1++=x%10+'0',x/=10; 84 while(s1--!=s)out(*s1); 85 } 86 void println(ll x){ 87 static char s[25],*s1;s1=s; 88 if (!x)*s1++='0';if (x<0)out('-'),x=-x; 89 while(x)*s1++=x%10+'0',x/=10; 90 while(s1--!=s)out(*s1); out('\n'); 91 } 92 void print(double x,int y){ 93 static ll mul[]={1,10,100,1000,10000,100000,1000000,10000000,100000000, 94 1000000000,10000000000LL,100000000000LL,1000000000000LL,10000000000000LL, 95 100000000000000LL,1000000000000000LL,10000000000000000LL,100000000000000000LL}; 96 if (x<-1e-12)out('-'),x=-x;x*=mul[y]; 97 ll x1=(ll)floor(x); if (x-floor(x)>=0.5)++x1; 98 ll x2=x1/mul[y],x3=x1-x2*mul[y]; print(x2); 99 if (y>0){out('.'); for (size_t i=1;i<y&&x3*mul[i]<mul[y];out('0'),++i) {}; print(x3);} 100 } 101 void println(double x,int y){print(x,y);out('\n');} 102 void print(char *s){while (*s)out(*s++);} 103 void println(char *s){while (*s)out(*s++);out('\n');} 104 void flush(){if (p1!=buf){fwrite(buf,1,p1-buf,stdout);p1=buf;}} 105 ~Ostream_fwrite(){flush();} 106 }Ostream; 107 inline void print(int x){Ostream.print(x);} 108 inline void println(int x){Ostream.println(x);} 109 inline void print(char x){Ostream.out(x);} 110 inline void println(char x){Ostream.out(x);Ostream.out('\n');} 111 inline void print(ll x){Ostream.print(x);} 112 inline void println(ll x){Ostream.println(x);} 113 inline void print(double x,int y){Ostream.print(x,y);} 114 inline void println(double x,int y){Ostream.println(x,y);} 115 inline void print(char *s){Ostream.print(s);} 116 inline void println(char *s){Ostream.println(s);} 117 inline void println(){Ostream.out('\n');} 118 inline void flush(){Ostream.flush();} 119 #undef ll 120 #undef OUT_SIZE 121 #undef BUF_SIZE 122 }; 123 using namespace IO; 124 125 126 typedef long long LL; 127 const int size=1e5+5; 128 int suma[size],sumb[size]; 129 struct frac{ 130 int son,mon; 131 frac(){} 132 frac(int _son,int _mon) 133 { 134 if(_son==0) {son=0,mon=1;} 135 else { 136 int g=__gcd(_son,_mon); 137 son=_son/g; 138 mon=_mon/g; 139 if(son>0&&mon<0)son=-son,mon=-mon; 140 } 141 } 142 friend bool operator<(frac x,frac y) 143 { 144 return 1LL*y.mon*x.son<1LL*x.mon*y.son; 145 } 146 friend bool operator==(frac x,frac y) 147 { 148 if((x.son==y.son)&&(x.mon==y.mon)) return true;; 149 return false; 150 } 151 friend bool operator<=(frac x,frac y) 152 { 153 if(x<y) return true; 154 if((x.son==y.son)&&(x.mon==y.mon)) return true; 155 return false; 156 } 157 }; 158 struct line{ 159 int a,b; 160 frac p; 161 line(){} 162 line(int a,int b):a(a),b(b),p(-b,a){} 163 friend bool operator<(line x,line y) 164 { 165 return x.p<y.p; 166 } 167 }L[size]; 168 vector<frac> ans; 169 int main() 170 { 171 int t; 172 //scanf("%d",&t); 173 read(t); 174 int n,c; 175 while(t--) 176 { 177 //scanf("%d%d",&n,&c); 178 read(n);read(c); 179 suma[0]=sumb[0]=0; 180 bool flag; 181 int a,b; 182 for(int i=1;i<=n;i++) 183 { 184 //scanf("%d%d",&a,&b); 185 read(a);read(b); 186 L[i]=line(a,b); 187 } 188 sort(L+1,L+1+n); 189 ans.clear(); 190 for(int i=1;i<=n;i++) 191 { 192 suma[i]=suma[i-1]+L[i].a; 193 sumb[i]=sumb[i-1]+L[i].b; 194 } 195 int newa=-suma[n],newb=-sumb[n]; 196 frac tp; 197 if(newa==0) 198 { 199 if(newb==c) 200 { 201 println(-1); 202 //puts("-1"); 203 continue; 204 } 205 } 206 if(newa!=0){ 207 tp=frac(c-newb,newa); 208 if(tp<=L[1].p) ans.push_back(tp); 209 } 210 flag=true; 211 for(int i=1;i<n;i++) 212 { 213 newa=2*suma[i]-suma[n]; 214 newb=2*sumb[i]-sumb[n]; 215 if(newa==0) 216 { 217 if(newb==c) 218 { 219 println(-1); 220 //puts("-1"); 221 flag=false; 222 break; 223 } 224 else continue; 225 } 226 tp=frac(c-newb,newa); 227 if(tp<=L[i+1].p&&L[i].p<=tp) ans.push_back(tp); 228 } 229 if(!flag) continue; 230 newa=suma[n],newb=sumb[n]; 231 if(newa==0) 232 { 233 if(newb==c) 234 { 235 println(-1); 236 //puts("-1"); 237 continue; 238 } 239 } 240 if(newa!=0){ 241 tp=frac(c-newb,newa); 242 if(L[n].p<=tp) ans.push_back(tp); 243 } 244 sort(ans.begin(),ans.end()); 245 ans.erase(unique(ans.begin(),ans.end()),ans.end()); 246 int len=ans.size(); 247 //printf("%d",len); 248 print(len); 249 for(int i=0;i<len;i++) 250 { 251 print(' '); 252 print(ans[i].son); 253 print('/'); 254 print(ans[i].mon); 255 //printf(" %d/%d",ans[i].son,ans[i].mon); 256 } 257 print('\n'); 258 //puts(""); 259 } 260 return 0; 261 }
1005 permutation 1
题解：https://blog.csdn.net/liufengwei1/article/details/98515203
1 #include<bits/stdc++.h> 2 #define maxl 45 3 using namespace std; 4 5 int n; 6 long long k; 7 int p[maxl]; 8 int ans[maxl]; 9 long long jc[maxl]; 10 int ok[maxl*2]; 11 struct node 12 { 13 int a[maxl]; 14 int in[maxl]; 15 void init() 16 { 17 for(int i=0;i<=20;i++) 18 a[i]=0,in[i]=false; 19 } 20 }; 21 vector <node> tmp; 22 vector <node> :: iterator it; 23 24 inline void prework() 25 { 26 scanf("%d%lld",&n,&k); 27 } 28 29 inline void solve(int u) 30 { 31 for(int i=1-n;i<=n-1;i++) 32 ok[i+n]=0; 33 bool flag;node d; 34 d.init(); 35 if(u==1) 36 { 37 tmp.clear(); 38 if(p[u]<0) 39 { 40 for(int i=n;i>=1-p[u];i--) 41 { 42 d.a[1]=i;d.a[2]=i+p[u]; 43 d.in[i]=true;d.in[i+p[u]]=true; 44 for(int j=1;j<=n;j++) 45 if(!d.in[j]) 46 ok[j-d.a[2]+n]++; 47 tmp.push_back(d); 48 d.in[i]=false;d.in[i+p[u]]=false; 49 } 50 } 51 else 52 { 53 for(int i=1;i<=n-p[u];i++) 54 { 55 d.a[1]=i;d.a[2]=i+p[u]; 56 d.in[i]=true;d.in[i+p[u]]=true; 57 for(int j=1;j<=n;j++) 58 if(!d.in[j]) 59 ok[j-d.a[2]+n]++; 60 tmp.push_back(d); 61 d.in[i]=false;d.in[i+p[u]]=false; 62 } 63 } 64 } 65 else 66 { 67 int x; 68 for(it=tmp.begin();it!=tmp.end();) 69 { 70 x=it->a[u]+p[u]; 71 if(x<=n && x>0) 72 { if( !(it->in[x])) 73 { 74 it->a[u+1]=x; 75 it->in[x]=true; 76 for(int j=1;j<=n;j++) 77 if(!(it->in[j])) 78 ok[j-x+n]++; 79 ++it; 80 } 81 else 82 tmp.erase(it); 83 } 84 else 85 tmp.erase(it); 86 } 87 } 88 } 89 90 inline int abs(int x) 91 { 92 if(x<0) return -x; 93 else return x; 94 } 95 96 inline void dfs(int u,long long res) 97 { 98 long long t; 99 int sum; 100 for(int i=1-n;i<=n-1;i++) 101 if(ok[i+n]) 102 { 103 sum=ok[i+n]; 104 if(k<=sum*jc[n-1-u]) 105 { 106 p[u]=i; 107 solve(u); 108 if(u<n-1) 109 dfs(u+1,k); 110 return; 111 } 112 k-=sum*jc[n-1-u]; 113 } 114 } 115 116 inline void mainwork() 117 { 118 for(int i=1-n;i<=n-1;i++) 119 ok[i+n]=n-abs(i); 120 ok[0+n]=0; 121 tmp.clear(); 122 dfs(1,k); 123 } 124 125 inline void print() 126 { 127 it=tmp.begin(); 128 for(int i=1;i<=n;i++) 129 printf("%d%c",it->a[i],(i==n)?'\n':' '); 130 } 131 132 int main() 133 { 134 jc[0]=1;jc[1]=1; 135 for(int i=1;i<=20;i++) 136 jc[i]=jc[i-1]*i; 137 int t; 138 scanf("%d",&t); 139 for(int i=1;i<=t;i++) 140 { 141 prework(); 142 mainwork(); 143 print(); 144 } 145 return 0; 146 }
1006 string matching
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 using namespace std; 5 typedef long long ll; 6 const int maxn = 2e6 + 5; 7 char ss[maxn]; 8 ll Next[maxn], extend[maxn]; 9 #define fi first 10 #define se second 11 #define INF 0x3f3f3f3f 12 13 //预处理计算Next数组 14 void getNext(char str[]) 15 { 16 int i = 0, j, po, len = strlen(str); 17 Next[0] = len; //初始化next[0] 18 while (str[i] == str[i + 1] && i + 1 < len) 19 i++; 20 Next[1] = i; //计算next[1] 21 po = 1; //初始化po的位置 22 for (i = 2; i < len; i++) { 23 if (Next[i - po] + i < Next[po] + po) //第一种情况，可以直接得到next[i]的值 24 Next[i] = Next[i - po]; 25 else //第二种情况，要继续匹配才能得到next[i]的值 26 { 27 j = Next[po] + po - i; 28 if (j < 0) 29 j = 0; //如果i>po+Next[po],则要从头开始匹配 30 while (i + j < len && str[j] == str[j + i]) 31 j++; 32 Next[i] = j; 33 po = i; //更新po的位置 34 } 35 } 36 } 37 38 //计算extend数组 39 void EXKMP(char s1[], char s2[]) 40 { 41 int i = 0, j, po, len = strlen(s1), l2 = strlen(s2); 42 getNext(s2); 43 while (s1[i] == s2[i] && i < l2 && i < len) 44 i++; 45 extend[0] = i; 46 po = 0; 47 for (i = 1; i < len; i++) { 48 if (Next[i - po] + i < extend[po] + po) 49 extend[i] = Next[i - po]; 50 else 51 { 52 j = extend[po] + po - i; 53 if (j < 0) 54 j = 0; 55 while (i + j < len && j < l2 && s1[j + i] == s2[j]) 56 j++; 57 extend[i] = j; 58 po = i; 59 } 60 } 61 } 62 63 int t; 64 int main() 65 { 66 scanf("%d", &t); 67 while (t--) 68 { 69 scanf("%s", ss); 70 EXKMP(ss, ss); 71 int len = strlen(ss); 72 ll ans = 0; 73 for(int i=1;i<len;++i) 74 { 75 if(extend[i] + i == len) 76 ans += extend[i]; 77 else 78 ans += extend[i] + 1; 79 } 80 printf("%lld\n", ans); 81 } 82 return 0; 83 }
1007 permutation 2
1 #include<bits/stdc++.h> 2 using namespace std; 3 const int mod = 998244353; 4 const int size=1e5+5; 5 int dp[size]; 6 void init() 7 { 8 dp[0]=1;dp[1]=1;dp[2]=1;dp[3]=2; 9 for(int i=4;i<size;i++) 10 { 11 dp[i]=(dp[i-1]+dp[i-3])%mod; 12 } 13 } 14 int main() 15 { 16 int t; 17 int n,x,y; 18 scanf("%d",&t); 19 init(); 20 while(t--) 21 { 22 scanf("%d%d%d",&n,&x,&y); 23 int beg,ed; 24 if(x==1) beg=1;else beg=x+1; 25 if(y==n) ed=n;else ed=y-1; 26 if(ed-beg<0) 27 { 28 puts("0"); 29 continue; 30 } 31 printf("%d\n",dp[ed-beg]); 32 } 33 }
1008 line symmetric
题解：https://blog.csdn.net/liufengwei1/article/details/98664030
1 #include<bits/stdc++.h> 2 #define maxl 1010 3 #define eps 1e-9 4 using namespace std; 5 6 inline int sgn(double x) 7 { 8 if(x>-eps && x<eps) return 0; 9 if(x<0) return -1; 10 else return 1; 11 } 12 13 struct point 14 { 15 double x,y; 16 point(double a=0,double b=0) 17 { 18 x=a,y=b; 19 } 20 point operator - (const point &b)const 21 { 22 return point(x-b.x,y-b.y); 23 } 24 point operator + (const point &b)const 25 { 26 return point(x+b.x,y+b.y); 27 } 28 friend point operator / (point p,double t) 29 { 30 return point(p.x/t,p.y/t); 31 } 32 }; 33 inline double dot(point a,point b) 34 { 35 return a.x*b.x+a.y*b.y; 36 } 37 inline double det(point a,point b) 38 { 39 return a.x*b.y-a.y*b.x; 40 } 41 struct line 42 { 43 point s,e;double k; 44 line(point a=point(),point b=point()) 45 { 46 s=a,e=b; 47 k=atan2(e.y-s.y,e.x-s.x); 48 } 49 bool operator < (const line &b)const 50 { 51 return k<b.k; 52 } 53 }; 54 55 int n; 56 point a[maxl]; 57 line b[maxl]; 58 bool ans,flag; 59 60 inline void prework() 61 { 62 scanf("%d",&n); 63 for(int i=1;i<=n;i++) 64 scanf("%lf%lf",&a[i].x,&a[i].y); 65 } 66 67 inline void mainwork() 68 { 69 if(n<=4) 70 { 71 ans=true; 72 return; 73 } 74 /*if(n==6 && a[1].x==0 && a[1].y==12) 75 { 76 ans=false; 77 return; 78 }*/ 79 ans=false;point mid,midd; 80 int s,t,ss,tt,cnt,lasts,lastt; 81 bool lastflag; 82 for(int i=1;i<=n;i++) 83 { 84 s=i-1;t=i+1; 85 if(s<1) s=n; 86 if(t>n) t=1; 87 mid=a[s]+a[t];mid=mid/2; 88 ss=s;tt=t;cnt=0;lastflag=false; 89 if(sgn(dot(a[i]-mid,a[t]-a[s]))!=0) 90 cnt=1; 91 for(int j=1;j<=n/2-1;j++) 92 { 93 94 lasts=ss;lastt=tt; 95 ss--;tt--; 96 if(ss<1) ss=n; 97 if(tt>n) tt=1; 98 midd=a[ss]+a[tt];midd=midd/2; 99 if(lastflag) 100 { 101 if(sgn(dot(a[ss]-mid,a[s]-mid))<=0 && 102 sgn(dot(a[lasts]-mid,a[s]-mid))<=0) 103 { 104 cnt++; 105 break; 106 } 107 if(sgn(dot(a[tt]-mid,a[t]-mid))<=0 && 108 sgn(dot(a[lastt]-mid,a[t]-mid))<=0) 109 { 110 cnt++; 111 break; 112 } 113 } 114 lastflag=false; 115 if(sgn(dot(a[t]-a[s],mid-midd))!=0 || 116 sgn(dot(a[tt]-a[ss],mid-midd))!=0) 117 cnt++,lastflag=true; 118 119 if(sgn(dot(a[ss]-mid,a[lasts]-mid))<=0 && 120 sgn(dot(a[tt]-mid,a[lastt]-mid))<=0) 121 cnt++,lastflag=true; 122 } 123 if(cnt<=1) 124 { 125 ans=true; 126 return; 127 } 128 } 129 for(int i=1;i<=n && !ans;i++) 130 { 131 s=i;t=i+1; 132 if(s<1) s=n; 133 if(t>n) t=1; 134 ss=s;tt=t;lastflag=false; 135 mid=a[s]+a[t];mid=mid/2;cnt=0; 136 for(int j=1;j<=(n-1)/2;j++) 137 { 138 139 lasts=ss;lastt=tt; 140 ss--;tt++; 141 if(ss<1) ss=n; 142 if(tt>n) tt=1; 143 if(lastflag) 144 { 145 if(sgn(dot(a[ss]-mid,a[s]-mid))<=0 && 146 sgn(dot(a[lasts]-mid,a[s]-mid))<=0) 147 { 148 cnt++; 149 break; 150 } 151 if(sgn(dot(a[tt]-mid,a[t]-mid))<=0 && 152 sgn(dot(a[lastt]-mid,a[t]-mid))<=0) 153 { 154 cnt++; 155 break; 156 } 157 } 158 lastflag=false; 159 midd=a[ss]+a[tt];midd=midd/2; 160 if(sgn(dot(a[t]-a[s],mid-midd))!=0 || 161 sgn(dot(a[tt]-a[ss],mid-midd))!=0) 162 cnt++,lastflag=true; 163 164 if(sgn(dot(a[ss]-mid,a[lasts]-mid))<=0 && 165 sgn(dot(a[tt]-mid,a[lastt]-mid))<=0) 166 cnt++,lastflag=true; 167 } 168 if(cnt<=1) 169 { 170 ans=true; 171 return; 172 } 173 } 174 } 175 176 inline void print() 177 { 178 if(ans) 179 puts("Y"); 180 else 181 puts("N"); 182 } 183 184 int main() 185 { 186 int t; 187 scanf("%d",&t); 188 for(int i=1;i<=t;i++) 189 { 190 prework(); 191 mainwork(); 192 print(); 193 } 194 return 0; 195 }
1009 discrete logarithm problem
unsolved
1010 find hidden array
unsolved
