qq_46173805的博客

Sample Gamesoltuionf(x)=∑i=0∞Pixif(x)=\sum_{i=0}^{∞}P_ix^if(x)=∑i=0∞​Pi​xi ,( PiP_iPi​ 表示构成长度 >i>i>i 的序列的可能性)由于序列为不下降序列，例：1112233451 1 1 2 2 3 3 4 5111223345，Pi=∏∑cntp=ipkP_i=\prod_{\sum cnt_p=i} p^kPi​=∏∑cntp​=i​pkf(x)=∏j=1∞∑k=0ipjkxk=∏j=

solution∏k=0∞aixi2+bixi+1\prod\limits_{k=0}^{∞}a_ix_i^2+b_ix_i+1k=0∏∞​ai​xi2​+bi​xi​+1考虑生成函数：考虑生成函数：考虑生成函数：=>∑k=0∞aik2xk+bixixk+xk=>\sum\limits_{k=0}^{∞}a_ik^2x^k+b_ix_ix^k+x^k=>k=0∑∞​ai​k2xk+bi​xi​xk+xk=∑k=0∞aik2xk+∑k=0∞bixixk+∑k=0∞xk,(三项分别当作A

卡特兰数给定nnn个000和nnn个111，它们将按照某种排序成长度为2n2n2n的序列，求它们能排列成的所有序列中，满足任意前缀序列中000的个数都不少于111的序列有多少个？（即从X走到Y路线不穿越，y=x的方案数）|~~~~~~~~~~~~~~~~~~~.Y(n,n)||||||.X(0,0)|--------------------------------------------------------------->Answer=C2nn−C2nn−1=C2nnn+1A

Counting Sequencessolution序列[a1,a2,a3...,an],(ai∈[1,m])，问序列中ai为偶数的个数非奇有几种序列[a_1,a_2,a_3...,a_n],(a_i∈[1,m])，问序列中a_i为偶数的个数非奇有几种序列[a1​,a2​,a3​...,an​],(ai​∈[1,m])，问序列中ai​为偶数的个数非奇有几种考虑生成函数：考虑生成函数：考虑生成函数：T=(1+x22!+x44!+x66!+...)m2∗(1+x1!+x22!+x33!+...)m+1

HDU-4045code/*SiberianSquirrel*//*CuteKiloFish*/#include<bits/stdc++.h>using namespace std;const int MOD = 1000000007;int n, r, k, m;int C[2010][2010];int S2[2000][2000];void solve(long long ans = 0, long long res = 0) { for(int i =

A Famous Gamecode/*Siberian Squirrel*//*Cute KiloFish*/#include<bits/stdc++.h>#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)#define ACM_LOCALusing namespace std;typedef long long ll;const double PI = acos(-1);const double

A. Little Pony and Expected Maximumsolutionres=∑i=1min−(i−1)nmn=∑i=1m(im)n−(i−1m)nres=\sum\limits_{i=1}^{m}\frac{i^n-(i-1)^n}{m^n}=\sum\limits_{i=1}^{m}(\frac {i}{m})^n-(\frac{i-1}{m})^nres=i=1∑m​mnin−(i−1)n​=i=1∑m​(mi​)n−(mi−1​)ncode/*Siberian Squirr

Candysolution∑i=1ni∗C2n−in−i∗(pn+1qn−i+qn+1pn−i)\sum\limits_{i=1}^{n}i*C_{2n-i}^{n-i}*(p^{n+1}q^{n-i}+q^{n+1}p^{n-i})i=1∑n​i∗C2n−in−i​∗(pn+1qn−i+qn+1pn−i)因为n<2e5，pn+1，qn+1过小，因此中间运算取对数，变为因为n<2e5，p^{n+1}，q^{n+1}过小，因此中间运算取对数，变为因为n<2e5，pn+1，qn+1过小

C. Bargaininputoutputcode

D. Count the Arraysinputoutputcode

E. Two Round Dancesinput20output12164510040883200code//Siberian Squirrel#include<bits/stdc++.h>#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)//#define ACM_LOCALusing namespace std;typedef long long ll;const int N = 2e5

B. Alice and the List of Presentsinput1 3output7code//Siberian Squirrel#include<bits/stdc++.h>#define IO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)//#define ACM_LOCALusing namespace std;typedef long long ll;const int N = 2e5 +

C. New Year and Permutationinput2020 437122297output265955509code#include<bits/stdc++.h>using namespace std;typedef long long ll;const double PI = acos(-1);const int INF = 0x3f3f3f3f;const int N = 3e5+10;const int MOD =1e9 + 7;//int n,

D. Radio Towersinput200000output202370013code#include<bits/stdc++.h>using namespace std;typedef long long ll;const double PI = acos(-1);const int INF = 0x3f3f3f3f;const int N = 3e5+10;const int MOD = 998244353;//int n, m, f[N];ll temp

Two Arraysinput723 9ouput157557417code#include<bits/stdc++.h>using namespace std;typedef long long ll;const double PI = acos(-1);const int INF = 0x3f3f3f3f;const int N = 3e5+10;const int MOD = 1e9 + 7;//int n, m, f[N];ll temp;//ll q