2021牛客暑期多校训练营2 B-Cannon（组合+推式子）
B-Cannon
首先nnn個(gè)炮在一行操作一次的方案數(shù)為2(n?2)2(n-2)2(n?2):前面兩個(gè)炮只能向右吃，最后兩個(gè)跑只能向左吃，而其余的炮既可以向左也可以向右，于是有4+2(n?4)4+2(n-4)4+2(n?4)種
于是操作mmm次的操作排列的方案數(shù)為2m(n?2)(n?3)...(n?m?1)=2m(n?2)!(n?2?m)!2^m(n-2)(n-3)...(n-m-1)=2^m\frac {(n-2)!}{(n-2-m)!}2m(n?2)(n?3)...(n?m?1)=2m(n?2?m)!(n?2)!?
設(shè)n=a?2,m=b?2n=a-2,m=b-2n=a?2,m=b?2
首先選擇從kkk選擇iii次操作第一行，選擇k?ik-ik?i次操作第二行，而問題一還需要從kkk個(gè)位置選擇出iii個(gè)按序放置操作序列，而問題二必須強(qiáng)制第一行操作放置在第二行操作前面，由此有：
對(duì)于問題1即是求
∑0≤i≤k(ki){2in!(n?i)!}×{2k?im![m?(k?i)]!}=2k∑0≤i≤kk!i!(k?i)!?n!(n?i)!?m![m?(k?i)]!=2kk!∑0≤i≤k(ni)(mk?i)=2kk!(n+mk)\begin{aligned} &\sum_{0\leq i\leq k}\dbinom{k}{i}\{2^i\frac{n!}{(n-i)!}\}×\{2^{k-i}\frac{m!}{[m-(k-i)]!}\}\\ \\&=2^k\sum_{0\leq i\leq k}\frac{k!}{i!(k-i)!}·\frac{n!}{(n-i)!}·\frac{m!}{[m-(k-i)]!}\\ \\&=2^kk!\sum_{0\leq i\leq k} \dbinom{n}{i}\dbinom{m}{k-i}\\\\&=2^kk!\dbinom{n+m}{k} \end{aligned}?0≤i≤k∑?(ik?){2i(n?i)!n!?}×{2k?i[m?(k?i)]!m!?}=2k0≤i≤k∑?i!(k?i)!k!??(n?i)!n!??[m?(k?i)]!m!?=2kk!0≤i≤k∑?(in?)(k?im?)=2kk!(kn+m?)?
對(duì)于問題2即是求
∑0≤i≤k{2in!(n?i)!}×{2k?im![m?(k?i)]!}=2k∑0≤i≤kn!(n?i)!×m![m?(k?i)]!=2kn!m!(n+m?k)!∑0≤i≤k(n+m?k)!(n?i)![m?(k?i)]!=2kn!m!(n+m?k)!∑0≤i≤k(n+m?kn?i)=2kn!m!(n+m?k)!∑n?k≤i≤n(n+m?kj)\begin{aligned} &\sum_{0\leq i\leq k}\{2^i\frac{n!}{(n-i)!}\}×\{2^{k-i}\frac{m!}{[m-(k-i)]!}\}\\\\&=2^k\sum_{0\leq i\leq k}\frac{n!}{(n-i)!}×\frac{m!}{[m-(k-i)]!}\\\\&=2^k\frac{n!m!}{(n+m-k)!}\sum_{0\leq i\leq k}\frac{(n+m-k)!}{(n-i)![m-(k-i)]!}\\\\&=2^k\frac{n!m!}{(n+m-k)!}\sum_{0\leq i\leq k}\dbinom{n+m-k}{n-i}\\\\&=2^k\frac{n!m!}{(n+m-k)!}\sum_{n-k\leq i\leq n}\dbinom{n+m-k}{j} \end{aligned}?0≤i≤k∑?{2i(n?i)!n!?}×{2k?i[m?(k?i)]!m!?}=2k0≤i≤k∑?(n?i)!n!?×[m?(k?i)]!m!?=2k(n+m?k)!n!m!?0≤i≤k∑?(n?i)![m?(k?i)]!(n+m?k)!?=2k(n+m?k)!n!m!?0≤i≤k∑?(n?in+m?k?)=2k(n+m?k)!n!m!?n?k≤i≤n∑?(jn+m?k?)?
其中下面一項(xiàng)是組合數(shù)種的連續(xù)多項(xiàng)之和，考慮前綴和！
∑n?k≤j≤n(n+m?kj)=∑0≤j≤n(n+m?kj)?∑0≤j≤n?k?1(n+m?kj)=∑0≤j≤n(n+m?kj)?∑0≤j≤n?k?1(n+m?kn+m?k?j)=∑0≤j≤n(n+m?kj)?∑m+1≤j≤n+m?k(n+m?kj)=∑0≤j≤n(n+m?kj)?{∑0≤j≤n+m?k(n+m?kj)?∑0≤j≤m(n+m?kj)}=∑0≤j≤n(n+m?kj)+∑0≤j≤m(n+m?kj)?2n+m?k=sum(n+m?k)\begin{aligned} &\sum_{n-k\leq j\leq n}\dbinom{n+m-k}{j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\sum_{0\leq j\leq n-k-1}\dbinom{n+m-k}{j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\sum_{0\leq j\leq n-k-1}\dbinom{n+m-k}{n+m-k-j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\sum_{m+1\leq j\leq n+m-k}\dbinom{n+m-k}{j}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}-\{\sum_{0\leq j\leq n+m-k}\dbinom{n+m-k}{j}-\sum_{0\leq j\leq m}\dbinom{n+m-k}{j}\}\\\\&=\sum_{0\leq j\leq n}\dbinom{n+m-k}{j}+\sum_{0\leq j\leq m}\dbinom{n+m-k}{j}-2^{n+m-k}=\text{sum}(n+m-k) \end{aligned}?n?k≤j≤n∑?(jn+m?k?)=0≤j≤n∑?(jn+m?k?)?0≤j≤n?k?1∑?(jn+m?k?)=0≤j≤n∑?(jn+m?k?)?0≤j≤n?k?1∑?(n+m?k?jn+m?k?)=0≤j≤n∑?(jn+m?k?)?m+1≤j≤n+m?k∑?(jn+m?k?)=0≤j≤n∑?(jn+m?k?)?{0≤j≤n+m?k∑?(jn+m?k?)?0≤j≤m∑?(jn+m?k?)}=0≤j≤n∑?(jn+m?k?)+0≤j≤m∑?(jn+m?k?)?2n+m?k=sum(n+m?k)?
考慮求出Sn,m=∑i=0m(ni)S_{n,m}=\sum_{i=0}^{m}\dbinom{n}{i}Sn,m?=∑i=0m?(in?)的關(guān)于nnn的遞推式
Sn,m=∑i=0m(ni)=∑i=0m[(n?1i)+(n?1i?1)]=Sn?1,m+Sn?1,m?1\begin{aligned} S_{n,m}=\sum_{i=0}^{m}\dbinom{n}{i}&=\sum_{i=0}^{m}[\dbinom{n-1}{i}+\dbinom{n-1}{i-1}]\\&=S_{n-1,m}+S_{n-1,m-1} \end{aligned}Sn,m?=i=0∑m?(in?)?=i=0∑m?[(in?1?)+(i?1n?1?)]=Sn?1,m?+Sn?1,m?1??
而Sn?1,m=∑i=0m(n?1i)=Sn?1,m?1+(n?1m)\begin{aligned} S_{n-1,m}=\sum_{i=0}^{m}\dbinom{n-1}{i}&=S_{n-1,m-1}+\dbinom{n-1}{m} \end{aligned}Sn?1,m?=i=0∑m?(in?1?)?=Sn?1,m?1?+(mn?1?)?
帶入可得出
Sn,m=2Sn?1,m?(n?1m)S_{n,m}=2S_{n-1,m}-\dbinom{n-1}{m} Sn,m?=2Sn?1,m??(mn?1?)
由此可得出
sum(x)=∑0≤j≤n(xj)+∑0≤j≤m(xj)?2x=Sx,n+Sx,m?2x=2sum(x?1)?(x?1n)?(x?1m)\begin{aligned} \text{sum}(x)&=\sum_{0\leq j\leq n}\dbinom{x}{j}+\sum_{0\leq j\leq m}\dbinom{x}{j}-2^{x}\\\\&=S_{x,n}+S_{x,m}-2^x\\\\&=2\text{sum}(x-1)-\dbinom{x-1}{n}-\dbinom{x-1}{m} \end{aligned}sum(x)?=0≤j≤n∑?(jx?)+0≤j≤m∑?(jx?)?2x=Sx,n?+Sx,m??2x=2sum(x?1)?(nx?1?)?(mx?1?)?
Code
#include<bits/stdc++.h> using namespace std; using ll=long long; template <class T=int> T rd() {T res=0;T fg=1;char ch=getchar();while(!isdigit(ch)) {if(ch=='-') fg=-1;ch=getchar();}while( isdigit(ch)) res=(res<<1)+(res<<3)+(ch^48),ch=getchar();return res*fg; } const ll mod=1e9+9; ll qmi(ll a,ll b) {ll v=1;while(b){if(b&1) v=v*a%mod;b>>=1;a=a*a%mod;}return v; } int n,m; ll inv[10000010]; void solve1() {ll ans=1;ll Two=1,Fact=1;ll C=1;for(int i=1;i<=n+m;i++){Two=Two*2%mod;Fact=Fact*i%mod;C=C*(n+m-i+1)%mod*inv[i]%mod;ans^=((Two*Fact%mod*C%mod)+mod)%mod;}printf("%lld ",ans); } ll sum[10000010];void solve2() {ll Cn=1,Cm=1;sum[0]=1;for(int i=1;i<=min(n,m);i++) sum[i]=2*sum[i-1]%mod;for(int i=min(n,m)+1;i<=n+m;i++){sum[i]=2*sum[i-1]-(i>n?Cn:0)-(i>m?Cm:0);if(i>n) Cn=Cn*i%mod*inv[i-n]%mod;if(i>m) Cm=Cm*i%mod*inv[i-m]%mod;sum[i]=(sum[i]%mod+mod)%mod;}ll Factn=1,Factm=1,Two=1;for(int i=1;i<=n;i++) Factn=Factn*i%mod;for(int i=1;i<=m;i++) Factm=Factm*i%mod;ll Inv=1;for(int i=1;i<=n+m;i++) Inv=Inv*inv[i]%mod;ll ans=Two*Factn%mod*Factm%mod*Inv%mod*sum[n+m]%mod;for(int i=1;i<=n+m;i++){Two=Two*2%mod;Inv=Inv*(n+m-i+1)%mod;ans^=((Two*Factn%mod*Factm%mod*Inv%mod*sum[n+m-i]%mod)+mod)%mod;}printf("%lld\n",ans); } int main() {n=rd()-2,m=rd()-2;inv[1]=1;for(int i=2;i<=n+m;i++) inv[i]=(mod-mod/i*inv[mod%i]%mod)%mod;solve1();solve2();return 0; }看著公式寫代碼都寫了好長(zhǎng)時(shí)間，wtcl
創(chuàng)作挑戰(zhàn)賽新人創(chuàng)作獎(jiǎng)勵(lì)來咯，堅(jiān)持創(chuàng)作打卡瓜分現(xiàn)金大獎(jiǎng)總結(jié)
以上是生活随笔為你收集整理的2021牛客暑期多校训练营2 B-Cannon（组合+推式子）的全部?jī)?nèi)容，希望文章能夠幫你解決所遇到的問題。
- 上一篇： 十大最恐怖短篇鬼故事 短篇恐怖的鬼故事
- 下一篇： 如何选购葡萄酒 选购葡萄酒方法