MemSQL Start[c]UP 2.0 - Round 1 C. Magic Trick

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Codeforces MemSQL Start[c]UP 2.0 - Round 1 C. Magic Trick

首先，我们先假设有抽出的牌样式为A

则，抽到同样的牌（不是同样类型）的概率为 1 / N

则，抽到不同的牌的概率为 N-1 / N

此时抽到A类型的概率为，在原来的N*M张中去掉我们最先抽出的一张A，再从中抽出剩下的 M-1张A类牌

综上所述，答案为

1 / N + ( N-1 ) / N * ( M-1 ) / ( M*N -1 )

特判当N等于1的时候，答案是1

#include <algorithm>#include <iostream>#include <iomanip>#include <cstring>#include <climits>#include <complex>#include <fstream>#include <cassert>#include <cstdio>#include <bitset>#include <vector>#include <deque>#include <queue>#include <stack>#include <ctime>#include <set>#include <map>#include <cmath>#define eps 1e-9#define INF 0x3f3f3f3fusing namespace std;typedef long long ll;typedef long double ld;typedef pair<ll, ll> pll;typedef complex<ld> point;typedef pair<int, int> pii;typedef pair<pii, int> piii;template<class T>inline bool read(T &n){    T x = 0, tmp = 1; char c = getchar();    while((c < '0' || c > '9') && c != '-' && c != EOF) c = getchar();    if(c == EOF) return false;    if(c == '-') c = getchar(), tmp = -1;    while(c >= '0' && c <= '9') x *= 10, x += (c - '0'),c = getchar();    n = x*tmp;    return true;}template <class T>inline void write(T n){    if(n < 0)    {        putchar('-');        n = -n;    }    int len = 0,data[20];    while(n)    {        data[len++] = n%10;        n /= 10;    }    if(!len) data[len++] = 0;    while(len--) putchar(data[len]+48);}//-----------------------------------int main(){    int n,m;    double ans=0.0;    read(n),read(m);    if(n==1)    {        ans=1.0;    }    else    {        ans=1.0/(double)n+(double)(n-1)/(double)n*(double)(m-1)/(double)(m*n-1);    }    printf("%.16f",ans);}

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