XIV Open Cup named after E.V. Pankratiev. GP of SPb

【XIV Open Cup named after E.V. Pankratiev. GP of SPb】的更多相关文章

XIV Open Cup named after E.V. Pankratiev. GP of SPb

A. Bracket Expression 直接按题意模拟即可. 时间复杂度$O(n)$. #include<stdio.h> #include<algorithm> #include<math.h> #include<string.h> #include<string> #include<vector> #include<set> #include<map> #include<queue> #in…

XIV Open Cup named after E.V. Pankratiev. GP of America

A. Ancient Diplomacy 建图,同色点间边权为$0$,异色点间边权为$1$,则等价于找一个点使得到它最短路最长的点的最短路最小,Floyd即可. 时间复杂度$O(n^3)$. #include<cstdio> #include<algorithm> using namespace std; const int N=110,inf=100000000; int n,m,i,j,k,x,y,a[N],g[N][N]; int main(){ while(~scanf(&…

XIV Open Cup named after E.V. Pankratiev. GP of Europe

A. The Motorway 等价于找到最小和最大的$L$满足存在$S$使得$S+(i-1)L\leq a_i\leq S+i\times L$ 即 $S\leq\min((1-i)L+a_i)$ $S\geq\max(-i\times L+a_i)$ 求出上下凸壳的交点即可,因为斜率本身有序,故时间复杂度为$O(n)$. #include<cstdio> const int N=1000010,BUF=12000000; const double inf=1e30,eps=1e-9; ch…

XVII Open Cup named after E.V. Pankratiev. GP of SPb

A. Array Factory 将下标按前缀和排序,然后双指针,维护最大的右边界即可. #include<cstdio> #include<algorithm> using namespace std; typedef long long ll; const int N=200010; int n,i,j,anslen,ansl,ansr,mr,q[N]; ll a[N],lim; inline bool cmp(int x,int y){return a[x]<a[y];…

XVI Open Cup named after E.V. Pankratiev. GP of SPB

A. Bubbles 枚举两个点,求出垂直平分线与$x$轴的交点,答案=交点数+1. 时间复杂度$O(n^2\log n)$. #include<cstdio> #include<algorithm> #include<cmath> using namespace std; const double eps=1e-9; int sgn(double x){ if(x<-eps)return -1; if(x>eps)return 1; return 0; }…

XIII Open Cup named after E.V. Pankratiev. GP of SPb

A. Graph Coloring 答案为$1$很好判,为$2$只需要二分图染色,对于$3$,首先爆搜哪些边要染成第$3$种颜色,然后二分图染色判定即可. B. Decimal Fraction 枚举前缀,那么只需要求出后面部分的最小循环节即可,将串翻转之后进行KMP,循环节长度$=i-next[i]$. 时间复杂度$O(n)$. C. Teams of Equal Power 首先将球员按能力值从大到小排序,假设一队的队长能力值比二队队长高,那么显然一队队长只能是第一个人,枚举二队队长,然后看…

XVIII Open Cup named after E.V. Pankratiev. GP of SPb

contest Link A. Base i − 1 Notation solved by sdcgvhgj 238 求出a+b的2进制后从低位到高两位两位地转化为i-1进制 i-1进制的第2k位和第2k+1位只有唯一的填法使2进制的第k位不留遗憾地攀登到高位 B. Squaring a Bit solved by rdc 77 本地打表,交表. C. Chickens solved by rdc, 24 f[i][mask] 表示前 i 个物品匹配 mask 集合的方案数. D. Lights…