BZOJ4830 [Hnoi2017]抛硬币 【扩展Lucas】

题目链接

BZOJ4830

题解

当\(a = b\)时，我们把他们投掷硬币的结果表示成二进制，发现，当\(A\)输给\(B\)时，将二进制反转一下\(A\)就赢了\(B\)

还要除去平局的情况，最后答案就是

\[\frac{2^{a + b} - {a + b \choose a}}{2}
\]

当\(a \neq b\)时，有些状态可能翻转后还是\(A\)赢\(B\)，需要加上这部分

\[\begin{aligned}
\sum\limits_{i = 0}^{b} \sum\limits_{j = 1}^{a - b - 1}{b \choose i} {a \choose i + j}
&= \sum\limits_{j = 1}^{a - b - 1} \sum\limits_{i = 0}^{b} {b \choose b - i} {a \choose i + j} \\
&= \sum\limits_{j = 1}^{a - b - 1} {a + b \choose b + j} \\
&= \sum\limits_{j = b + 1}^{a - 1} {a + b \choose j} \\
\end{aligned}
\]

答案是

\[\frac{2^{a + b} + \sum\limits_{j = b + 1}^{a - 1} {a + b \choose j} }{2}
\]

除\(2\)的处理，因为组合数是对称的，所以只算一半

如果中间单独剩一个，一定可以被\(2\)整除，处理因子时减去一个即可

由于要模\(10^{K}\)，组合数的计算用扩展\(Lucas\)

此题非常卡常，要使用扩展\(Lucas\)的一些优化

1.预处理阶乘

2.当\(p\)的幂次大于\(k\)时直接返回\(0\)

3.没了

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<iomanip>
#include<cstdio>
#include<vector>
#include<queue>
#include<ctime>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (LL i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 2000005,maxm = 100005,INF = 0x3f3f3f3f;
int K,pr[2],pk[2],P,fac[2][maxn],now,ans;
LL A,B;
void init(){
	pr[0] = 2; pr[1] = 5; pk[0] = pk[1] = P = fac[0][0] = fac[1][0] = 1;
	REP(i,K) pk[0] *= 2,pk[1] *= 5,P *= 10;
	for (LL i = 1; i < pk[0]; i++)
		if (i % 2) fac[0][i] = 1ll * fac[0][i - 1] * i % pk[0];
		else fac[0][i] = fac[0][i - 1];
	for (LL i = 1; i < pk[1]; i++)
		if (i % 5) fac[1][i] = 1ll * fac[1][i - 1] * i % pk[1];
		else fac[1][i] = fac[1][i - 1];
}
inline int qpow(int a,LL b,int p){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % p)
		if (b & 1) re = 1ll * re * a % p;
	return re;
}
inline void exgcd(int a,int b,int&d ,int& x,int& y){
	if (!b){d = a; x = 1; y = 0;}
	else exgcd(b,a % b,d,y,x),y -= (a / b) * x;
}
inline int inv(int n,int p){
	int d,x,y; exgcd(n,p,d,x,y);
	return (x % p + p) % p;
}
int Fac(LL n,int pk,int p){
	if (!n) return 1;
	return 1ll * qpow(fac[now][pk - 1],n / pk,pk) * fac[now][n % pk] % pk * Fac(n / p,pk,p) % pk;
}
int C(LL n,LL m,int pk,int p,bool f){
	LL k = 0;
	for (LL i = n; i; i /= p) k += i / p;
	for (LL i = m; i; i /= p) k -= i / p;
	for (LL i = n - m; i; i /= p) k -= i / p;
	if (p == 2 && f) k--;
	if (k >= 9) return 0;
	now = (p == 5);
	LL a = Fac(n,pk,p),b = Fac(m,pk,p),c = Fac(n - m,pk,p),ans;
	ans = a * inv(b,pk) % pk * inv(c,pk) % pk;
	if (p == 5 && f) ans = 1ll * ans * inv(2,pk) % P;
	ans = ans * qpow(p,k,pk) % pk;
	return ans * (P / pk) % P * inv(P / pk,pk) % P;
}
int exlucas(LL n,LL m,bool f){
	if (m > n) return 0;
	int re = 0;
	re = (re + C(n,m,pk[0],pr[0],f)) % P;
	re = (re + C(n,m,pk[1],pr[1],f)) % P;
	return re;
}
int main(){
	//double t = clock();
	K = 9; init();
	while (~scanf("%lld%lld%d",&A,&B,&K)){
		ans = qpow(2,A + B - 1,P);
		if (A == B) ans = ((ans - exlucas(A + B,A,1)) % P + P) % P;
		else {
			for (LL i = ((A + B) >> 1) + 1; i < A; i++){
				ans = (ans + exlucas(A + B,i,0)) % P;
			}
			if ((A + B) % 2 == 0) ans = (ans + exlucas(A + B,(A + B) >> 1,1)) % P;
		}
		int md = qpow(10,K,INF); ans %= md;
		while (ans < md / 10) putchar('0'),md /= 10;
		printf("%d\n",ans);
	}
	//cerr << (clock() - t) / CLOCKS_PER_SEC << endl;
	return 0;
}