XVI Open Cup named after E.V. Pankratiev. GP of Ekaterinburg.
贼惨 130/186
B Black Widow
简单题
#include <bits/stdc++.h> const long long mod = 1e9+7; const double ex = 1e-10; #define inf 0x3f3f3f3f using namespace std; map <int,int> M; int a[1010]; int b[100010]; int main() { int N; scanf("%d",&N); int t = 0; for (int i = 1; i<=N;i++){ scanf("%d",&a[i]); t = max(t,a[i]); for (int j = 1; j*j<=a[i];j++) if (a[i]%j==0) { b[j] = 1; if (a[i]/j > 100000)M[a[i]/j]++; else b[a[i]/j] = 1; } } for (int i = 2 ;i<=1000000001 ;i++) if ((i<=100000&&b[i]==0)||(i>100000&&M[i]==0)) { printf("%d\n",i); return 0; } return 0; }
C Chitauri
海盗分金，按照海盗分金的贪心策略递推的求。
#include <bits/stdc++.h> using namespace std; int n,k; int vis[3005],money[3005],last; int main() { scanf("%d%d",&n,&k); money[1] = k; last = 1; for (int i=2;i<=n;++i) { int cnt = 0; int gg = 0; for (int j = 1;j<=i-1;j++){ if (money[j] == 0) cnt++; if (money[j] == -1) gg++; } if (2*(min(cnt,k)+gg+1) < i ) { money[i] = -1; continue; } last = i; int tk=k; if (k-((i+1)/2 - gg -1) > 0) money[i] =k-((i+1)/2 - gg -1),tk-=money[i]; for (int j=i-1;j>=1;--j) { if (money[j]==-1) money[j]=0; else if (money[j]==0) { if (tk>0) { money[j]=1; --tk; } } else money[j]=0; } } for (int z=n;z>=1;--z) printf("%d%c",money[z],z==1?'\n':' '); return 0; }
D Dr. Banner
简单DP
#include <bits/stdc++.h> using namespace std; const long long mod=1e9+7; int n; long long dp[21][100005]; int main() { scanf("%d",&n); dp[0][n]=1; for (int i=1;i<=20;++i) { for (int j=1;j<=n;++j) dp[i-1][j]=(dp[i-1][j]+dp[i-1][j-1])%mod; for (int j=1;j*2<=n;++j) dp[i][j]=(dp[i-1][n]-dp[i-1][2*j-1]+mod)%mod; } long long res=0; for (int i=0;i<=20;++i) res=(res+dp[i][n])%mod; cout<<res<<endl; return 0; }
E Egocentric Loki (补)
题目很长，其实就是枚举每个三角形，判断是否有点处于这个三角形和它的外接圆之间。
精度的要求还是比较低的，就是输入数据范围有误，，心态崩了。
#include <bits/stdc++.h> using namespace std; const long double eps=1e-8; int sgn ( long double x ) { return ( x > eps ) - ( x < -eps ) ; } struct Point{ long double x,y; Point(){} Point(long double _x,long double _y){ x=_x;y=_y; } Point operator +(const Point &b)const{ return Point(x+b.x,y+b.y); } Point operator /(const long double &k)const{ return Point(x/k,y/k); } Point operator -(const Point &b)const { return Point(x-b.x,y-b.y); } Point operator *(const long double &rhs)const{ return Point(x*rhs,y*rhs); } long double operator ^(const Point &b) const { return x*b.y-y*b.x; } long double distance(Point p) { return (x-p.x)*(x-p.x)+(y-p.y)*(y-p.y); } Point rotleft(){ return Point(-y,x); } }; struct Line{ Point s,e; Line(){} Line(Point _s,Point _e){ s=_s; e=_e; } /* Point line_intersection (Line v) {//( P a , P b , P p , P q ) { double U = cross ( p - a , q - p ) ; double D = cross ( b - a , q - p ) ; return a + ( b - a ) * ( U / D ) ; } */ Point crosspoint(Line v) { long double U=(v.s-s)^(v.e-v.s); long double D=(e-s)^(v.e-v.s); return s+(e-s)*(U/D); } }; struct Circle{ Point p;long double r; Circle(){} Circle(Point a,Point b,Point c){ Line u=Line((a+b)/2.0,(a+b)/2.0+(a-b).rotleft()); Line v=Line((b+c)/2.0,(b+c)/2.0+(c-b).rotleft()); p=u.crosspoint(v); r=p.distance(a); } int relation(Point b){ long double dst=b.distance(p); if(sgn(dst-r)<0) return 2; else if(sgn(dst-r)==0) return 1; return 0; } }; inline double area(double x1,double y1,double x2,double y2,double x,double y) { double X1=x1-x,Y1=y1-y,X2=x2-x,Y2=y2-y; return fabs(X1*Y2-X2*Y1); } inline bool intriangle(double x1,double y1,double x2,double y2,double x3,double y3,double x,double y) { int flag=sgn(area(x1,y1,x2,y2,x3,y3)-area(x1,y1,x2,y2,x,y)-area(x1,y1,x3,y3,x,y)-area(x2,y2,x3,y3,x,y)); return flag==0; } int n; long double xx1[10010],yy1[10010],xx2[10010],yy2[10010],xx3[10010],yy3[10010]; int main() { while (scanf("%d",&n)==1) { for(int i=1;i<=n;i++) cin>>xx1[i]>>yy1[i]>>xx2[i]>>yy2[i]>>xx3[i]>>yy3[i]; int flag=0; for(int i=1;i<=n;i++) { Circle now=Circle(Point(xx1[i],yy1[i]),Point(xx2[i],yy2[i]),Point(xx3[i],yy3[i])); //cout<<now.r<<" "<<now.p.x<<" "<<now.p.y<<endl; for(int j=1;j<=n;j++) { if(j==i) continue; // cout<<now.relation(Point(x1[j],y1[j]))<<endl; if(!intriangle(xx1[i],yy1[i],xx2[i],yy2[i],xx3[i],yy3[i],xx1[j],yy1[j])&&(now.relation(Point(xx1[j],yy1[j]))>=1)) flag=1; if(!intriangle(xx1[i],yy1[i],xx2[i],yy2[i],xx3[i],yy3[i],xx2[j],yy2[j])&&(now.relation(Point(xx2[j],yy2[j]))>=1)) flag=1; if(!intriangle(xx1[i],yy1[i],xx2[i],yy2[i],xx3[i],yy3[i],xx3[j],yy3[j])&&(now.relation(Point(xx3[j],yy3[j]))>=1)) flag=1; if(flag) break; } if(flag) break; } if(flag) printf("NO\n"); else printf("YES\n"); } return 0; }
F Fury
缩点 ，然后DAG上贪心
#include <bits/stdc++.h> const long long mod = 1e9+7; const double ex = 1e-10; #define REP(i,x,y) for(int i=x;i<(y);i++) #define RREP(i,x,y) for(int i=x;i>(y);i--) #define inf 0x3f3f3f3f #define maxn 310 using namespace std; int n,m,pre[maxn],low[maxn],sccno[maxn],dfs_clock,scc_cnt; vector<int>e[maxn]; stack<int>S; void dfs(int st) { pre[st]=low[st]=(++dfs_clock); S.push(st); REP(i,0,e[st].size()) { int nxt=e[st][i]; if(!pre[nxt]) { dfs(nxt); low[st]=min(low[st],low[nxt]); } else if(!sccno[nxt]) { low[st]=min(low[st],pre[nxt]); } } if(low[st]==pre[st]) { scc_cnt++; while(1) { int now=S.top();S.pop(); sccno[now]=scc_cnt; if(now==st) break; } } } void Tarjan() { dfs_clock=scc_cnt=0; memset(sccno,0,sizeof(sccno)); memset(pre,0,sizeof(pre)); for(int i=1;i<=n;i++) if(!pre[i]) dfs(i); } struct edge{ int v,f,u,vv; edge(){} edge(int _v,int _f,int u,int v):v(_v),f(_f),u(u),vv(v){} }; vector<edge>DAG[maxn]; int col[maxn],DAG_cnt; void gao() { REP(i,1,n+1) { int u=sccno[i]; REP(j,0,e[i].size()) { int v=sccno[e[i][j]]; if(v==u) continue; DAG_cnt++; DAG[u].push_back(edge(v,0,i,e[i][j])); } } } int num[maxn]; int dfs2(int rt) { num[rt]++; for(int i=0;i<DAG[rt].size();i++) { int nxt=DAG[rt][i].v; if(DAG[rt][i].f==1) continue; if(num[nxt]>=1) {num[nxt]++;continue;} dfs2(nxt); } } vector<int>SCC[maxn]; typedef pair<int,int> pii; vector<pii>Ans; int main() { scanf("%d %d",&n,&m); REP(i,1,m+1) { int u,v;scanf("%d %d",&u,&v); e[u].push_back(v); } Tarjan(); int ans=0; for(int i=1;i<=n;i++) SCC[sccno[i]].push_back(i); for(int i=1;i<=scc_cnt;i++){ if(SCC[i].size()==1) continue; for(int j=0;j<SCC[i].size();j++){ Ans.push_back(make_pair(SCC[i][j],SCC[i][(j+1)%SCC[i].size()])); } } // REP(i,0,Ans.size()) cout<<Ans[i].first<<" "<<Ans[i].second<<endl; for(int i=1;i<=n;i++) col[sccno[i]]++; for(int i=1;i<=scc_cnt;i++) if(col[i]>1) ans+=col[i]; gao(); for(int i=1;i<=scc_cnt;i++) { for(int i=1;i<=scc_cnt;i++) num[i]=0; dfs2(i); for(int j=0;j<DAG[i].size();j++) { if(num[DAG[i][j].v]<=1) continue; DAG_cnt--; num[DAG[i][j].v]--; DAG[i][j].f=1; } for(int j=0;j<DAG[i].size();j++) { if(DAG[i][j].f==1) continue; Ans.push_back(make_pair(DAG[i][j].u,DAG[i][j].vv)); } } /* for(int i=1;i<=scc_cnt;i++) { printf("%d: %d\n",i,DAG[i].size()); for(int j=0;j<DAG[i].size();j++) { printf("%d ",DAG[i][j].v); } puts(""); }*/ printf("%d %d\n",n,DAG_cnt+ans); REP(i,0,Ans.size()) { printf("%d %d\n",Ans[i].first,Ans[i].second); } return 0; }
G Groot
签到
#include <bits/stdc++.h> const long long mod = 1e9+7; const double ex = 1e-10; #define inf 0x3f3f3f3f using namespace std; int main() { string s; cin >> s ; cin >> s ; cin >> s ; if (s.length() == 5) puts("Pfff"); else { cout <<'W'; for (int i = 1; i<=s.length()-5;i++) cout <<'o'; puts("w"); } return 0; }
J 待补
