题目链接
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 eq b)时，有些状态可能翻转后还是(A)赢(B)，需要加上这部分
[egin{aligned}
sumlimits_{i = 0}^{b} sumlimits_{j = 1}^{a - b - 1}{b choose i} {a choose i + j}
&= sumlimits_{j = 1}^{a - b - 1} sumlimits_{i = 0}^{b} {b choose b - i} {a choose i + j} \
&= sumlimits_{j = 1}^{a - b - 1} {a + b choose b + j} \
&= sumlimits_{j = b + 1}^{a - 1} {a + b choose j} \
end{aligned}
]
答案是
[frac{2^{a + b} + sumlimits_{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
",ans);
}
//cerr << (clock() - t) / CLOCKS_PER_SEC << endl;
return 0;
}