扔硬币
题解：
1.如果(left ( m+k ight )> n)，那么就很明显答案为0；
2.根据条件概率：则题目就是求，在至少有(m)枚硬币是反面的情况下，恰好有(k)枚硬币是正面的概率。那么就可以设(A)为至少有(m)枚硬币是反面，(B)为恰好有(k)枚硬币是正面，答案即为(frac{Pleft ( AB ight )}{Pleft ( A ight )})；恰好(k)枚硬币是正面，则(Pleft ( AB ight )=C_{n}^{k})，至少(m)枚硬币是反面，则(Pleft ( A ight )=left ( C_{n}^{0}+C_{n}^{1}+cdots +C_{n}^{n} ight )-left ( C_{n}^{0}+C_{n}^{1}+cdots +C_{n}^{m-1} ight )=2^{n}-sum_{i=0}^{m-1}C_{n}^{i})，所以答案即为：(C_{n}^{k}/left ( 2^{n}-sum_{i=0}^{m-1}C_{n}^{i} ight )(除法又可以转为求逆元))
AC_Code:
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5+10; 5 const ll mod=1e9+7; 6 7 ll n,m,k; 8 ll F[maxn]; 9 ll qpow(ll a, ll b){//快速幂 10 ll res=1; 11 while( b ){ 12 if( b&1 ) res = res*a%mod; 13 a = a*a%mod; 14 b>>=1; 15 } 16 return res; 17 } 18 19 ll inv(ll x){//求逆元 20 return qpow(x,mod-2); 21 } 22 23 void init(){ 24 F[0]=1; 25 for(int i=1;i<=100000;i++){ 26 F[i]=F[i-1]*i%mod; 27 } 28 } 29 30 int main() 31 { 32 int t; 33 scanf("%d",&t); 34 init(); 35 while( t-- ){ 36 scanf("%lld%lld%lld",&n,&m,&k); 37 if( m+k>n ){ 38 printf("0 "); 39 continue; 40 } 41 ///C(n,k):n!/(k!*(n-k)!) 42 ll up=F[n]*inv(F[k])%mod*inv(F[n-k])%mod; 43 //2^n-sum(C(n,0)~C(n,m-1)); 44 ll down=qpow(2,n); 45 ll sum=1,pre=1; 46 for(ll i=1;i<m;i++){ 47 pre=pre*(n-i+1)%mod*inv(i)%mod; 48 sum = (sum+pre)%mod; 49 } 50 down = (down-sum+mod)%mod; 51 down = inv(down); 52 ll ans=up*down%mod; 53 printf("%lld ",ans); 54 } 55 return 0; 56 }
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 const int maxn = 1e5+10; 5 const ll mod=1e9+7; 6 7 ll n,m,k; 8 ll F[maxn];//F[i]:1~i的阶乘 9 ll Inv[maxn];//Inv[i]:i的逆元 10 ll Finv[maxn];//Finv[i]:F[i]的逆元 11 12 ll qpow(ll a, ll b){//快速幂 13 ll res=1; 14 while( b ){ 15 if( b&1 ) res = res*a%mod; 16 a = a*a%mod; 17 b>>=1; 18 } 19 return res; 20 } 21 22 ll inv(ll x){//求逆元 23 return qpow(x,mod-2); 24 } 25 26 void init(){ 27 F[0]=F[1]=1; 28 Finv[0]=Finv[1]=1; 29 Inv[0]=Inv[1]=1; 30 for(int i=2;i<=100000;i++){ 31 F[i]=1ll*F[i-1]*i%mod; 32 Inv[i]=1ll*(mod-mod/i)*Inv[mod%i]%mod; 33 Finv[i]=1ll*Finv[i-1]*Inv[i]%mod; 34 } 35 } 36 37 ll C(ll a,ll b){//求C(a,b) 38 if( b>a || b<0 ) return 0ll; 39 return 1ll*F[a]*Finv[b]%mod*Finv[a-b]%mod; 40 } 41 42 int main() 43 { 44 int t; 45 scanf("%d",&t); 46 init(); 47 while( t-- ){ 48 scanf("%lld%lld%lld",&n,&m,&k); 49 if( m+k>n ){ 50 printf("0 "); 51 continue; 52 } 53 ll up=C(n,k);//分母 54 ll down=qpow(2,n); 55 ll sum=1; 56 for(int i=1;i<m;i++){ 57 sum=(sum+C(n,i))%mod; 58 } 59 down=(down-sum+mod)%mod;//分子 60 ll ans=up*inv(down)%mod; 61 printf("%lld ",ans); 62 63 } 64 return 0; 65 }