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 }
时间： 2024-10-03 13:29:59