codeforces gym 100357 H (DP 高精度）

Posted 2020-09-16 rpSebastian

题目大意

　　有r*s张扑克牌，数字从1到 r，每种数字有s种颜色。

　　询问对于所有随机的d张牌，能选出c张组成顺子的概率和组成同花的概率。

解题分析

　　对于组成顺子的概率，令dp[i][j][k]表示一共选出了i张牌，数字从1~j，最后有k张牌是顺子。对于每个数字进行考虑，有0~s种选法。要保证连续c张牌的顺子。

　　对于组成同花的概率，令dp[i][j]表示一共选出了i张牌，颜色从1~j，。对于每种颜色进行考虑，有0~r种选法。要保证没有c张牌是相同颜色的。

　　最后用高精度来输出答案。

参考程序

  1 #include <bits/stdc++.h>
  2 using namespace std;
  3 
  4 const int N=30;
  5 
  6 class bign  
  7 {  
  8     public:
  9     enum {MAXN = 100};
 10     int len, s[MAXN];  
 11     void clean()  
 12     {  
 13         while(len > 1 && !s[len-1]) len--;  
 14     }
 15     bign ()  
 16     {  
 17         memset(s, 0, sizeof(s));  
 18         len = 1;  
 19     }  
 20     bign (int num) { *this = num; }  
 21     bign (long long num) { *this = num; }  
 22     bign (const char *num) { *this = num; }
 23     bign operator = (const long long num)  
 24     {  
 25         char s[MAXN];  
 26         sprintf(s, "%I64d", num);  
 27         *this = s;  
 28         return *this;  
 29     }    
 30     bign operator = (const int num)  
 31     {  
 32         char s[MAXN];  
 33         sprintf(s, "%d", num);  
 34         *this = s;  
 35         return *this;  
 36     }  
 37     bign operator = (const char *num)  
 38     {  
 39         for(int i = 0; num[i] == ‘0‘ && num[1]!=‘\0‘; num++) ;  //去前导0  
 40         len = strlen(num);  
 41         for(int i = 0; i < len; i++) s[i] = num[len-i-1] - ‘0‘;  
 42         return *this;  
 43     }  
 44     bign operator + (const bign &b) const //+  
 45     {  
 46         bign c;  
 47         c.len = 0;  
 48         for(int i = 0, g = 0; g || i < max(len, b.len); i++)  
 49         {  
 50             int x = g;  
 51             if(i < len) x += s[i];  
 52             if(i < b.len) x += b.s[i];  
 53             c.s[c.len++] = x % 10;  
 54             g = x / 10;  
 55         }  
 56         return c;  
 57     }  
 58     bign operator += (const bign &b)  
 59     {  
 60         *this = *this + b;  
 61         return *this;  
 62     }    
 63     bign operator * (const int x)
 64     {
 65         bign c;  
 66         int j=0; for (int y=x;y;y/=10,j++);
 67         c.len = len + j;          
 68         for(int i = 0; i < len; i++)  
 69             c.s[i]  = s[i] * x;
 70 
 71         for(int i = 0; i < c.len; i++)  
 72         {  
 73             c.s[i+1] += c.s[i]/10;  
 74             c.s[i] %= 10;  
 75         }
 76         c.clean();  
 77         return c;  
 78     }
 79     bign operator * (const bign &b) //*  
 80     {  
 81         bign c;  
 82         c.len = len + b.len;  
 83         for(int i = 0; i < len; i++)  
 84         {  
 85             for(int j = 0; j < b.len; j++)  
 86             {  
 87                 c.s[i+j] += s[i] * b.s[j];  
 88             }  
 89         }  
 90         for(int i = 0; i < c.len; i++)  
 91         {  
 92             c.s[i+1] += c.s[i]/10;  
 93             c.s[i] %= 10;  
 94         }  
 95         c.clean();  
 96         return c;  
 97     }  
 98     bign operator *= (const bign &b)  
 99     {  
100         *this = *this * b;  
101         return *this;  
102     }    
103     bign operator *= (const int x)  
104     {  
105         *this = *this * x;  
106         return *this;  
107     }  
108     bign operator - (const bign &b)  
109     {  
110         bign c;  
111         c.len = 0;  
112         for(int i = 0, g = 0; i < len; i++)  
113         {  
114             int x = s[i] - g;  
115             if(i < b.len) x -= b.s[i];  
116             if(x >= 0) g = 0;  
117             else  
118             {  
119                 g = 1;  
120                 x += 10;  
121             }  
122             c.s[c.len++] = x;  
123         }  
124         c.clean();  
125         return c;  
126     }  
127     bign operator -= (const bign &b)  
128     {  
129         *this = *this - b;  
130         return *this;  
131     }  
132     bign operator / (const bign &b)  
133     {  
134         bign c, f = 0;  
135         for(int i = len-1; i >= 0; i--)  
136         {  
137             f = f*10;  
138             f.s[0] = s[i];  
139             while(f >= b)  
140             {  
141                 f -= b;  
142                 c.s[i]++;  
143             }  
144         }  
145         c.len = len;  
146         c.clean();  
147         return c;  
148     }  
149     bign operator /= (const bign &b)  
150     {  
151         *this  = *this / b;  
152         return *this;  
153     }  
154     bign operator % (const bign &b)  
155     {  
156         bign r = *this / b;  
157         r = *this - r*b;  
158         return r;  
159     }  
160     bign operator %= (const bign &b)  
161     {  
162         *this = *this % b;  
163         return *this;  
164     }  
165     bool operator < (const bign &b)  
166     {  
167         if(len != b.len) return len < b.len;  
168         for(int i = len-1; i >= 0; i--)  
169         {  
170             if(s[i] != b.s[i]) return s[i] < b.s[i];  
171         }  
172         return false;  
173     }  
174     bool operator > (const bign &b)  
175     {  
176         if(len != b.len) return len > b.len;  
177         for(int i = len-1; i >= 0; i--)  
178         {  
179             if(s[i] != b.s[i]) return s[i] > b.s[i];  
180         }  
181         return false;  
182     }  
183     bool operator == (const bign &b)  
184     {  
185         return !(*this > b) && !(*this < b);  
186     }  
187     bool operator != (const bign &b)  
188     {  
189         return !(*this == b);  
190     }  
191     bool operator <= (const bign &b)  
192     {  
193         return *this < b || *this == b;  
194     }  
195     bool operator >= (const bign &b)  
196     {  
197         return *this > b || *this == b;  
198     }  
199     operator string() const  
200     {  
201         string res = "";  
202         for(int i = 0; i < len; i++) res = char(s[i]+‘0‘) + res;  
203         return res;  
204     }  
205     friend istream& operator >> (istream &in, bign &x)  
206     {  
207         string s;  
208         in >> s;  
209         x = s.c_str();  
210         return in;  
211     } 
212     friend ostream& operator << (ostream &out, const bign &x)  
213     {  
214         out << string(x);
215         return out;  
216     } 
217 };
218 
219 int r,s,d,c;
220 int C[N][N];
221 bign dp[N][N];
222 bign f[N][N][N];
223 
224 bign calc_1(int n,int k)
225 {
226     bign tmp=1;
227     for (int i=1;i<=k;i++) 
228     {
229         tmp=tmp*(n-(i-1));
230         tmp=tmp/i;
231     }
232     return tmp;
233 }
234 
235 bign gcd(bign x,bign y)
236 {
237     //return y>0?gcd(y,x % y):x;
238     bign tmp;
239     if (y>0) tmp=gcd(y,x % y); else tmp=x;
240     return tmp;
241 }
242 
243 int main()
244 {
245     freopen("poker.in","r",stdin);
246     freopen("poker.out","w",stdout);
247 
248     for (int i=0;i<N;i++) C[i][0]=1;
249     for (int i=1;i<N;i++)
250         for (int j=1;j<=i;j++)
251             C[i][j]=C[i-1][j]+C[i-1][j-1];
252         
253     cin.sync_with_stdio(0);
254     while (cin>>r>>s>>d>>c)
255     {
256         if (!r && !s && !d && !c) break;
257 
258         memset(f,0,sizeof(f));
259         for (int j=0;j<=r;j++) f[0][j][0]=1;
260         for (int i=1;i<=d;i++)
261             for (int j=1;j<=r;j++)
262             {
263                 for (int k=0;k<=min(i,c-1);k++) 
264                     f[i][j][0]+=f[i][j-1][k];
265                 for (int k=1;k<=min(j,c-1);k++)
266                     for (int l=1;l<=min(i,s);l++)
267                         f[i][j][k]+=f[i-l][j-1][k-1]*C[s][l];
268             }
269         bign ans=0;
270         for (int k=0;k<=c-1;k++) ans+=f[d][r][k];
271         bign a=calc_1(r*s,d),b=a-ans;
272         cout<<b/gcd(a,b)<<"/"<<a/gcd(a,b)<<endl;
273 
274         memset(dp,0,sizeof(dp));
275         for (int j=0;j<=s;j++) dp[0][j]=1;
276         for (int i=1;i<=d;i++)
277             for (int j=1;j<=s;j++)
278                 for (int k=0;k<=min(i,c-1);k++)
279                     dp[i][j]+=dp[i-k][j-1]*C[r][k];
280         a=calc_1(r*s,d),b=a-dp[d][s];
281         cout<<b/gcd(a,b)<<"/"<<a/gcd(a,b)<<endl<<endl;
282     }
283 }

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