Who’s Aunt Zhang
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 19 Accepted Submission(s): 16
Problem Description Aunt Zhang, well known as 张阿姨, is a fan of Rubik’s cube. One day she buys a new one and would like to color it as a gift to send to Teacher Liu, well known as 刘老师. As Aunt Zhang is so ingenuity, she can color all the cube’s points, edges and faces with K different color. Now Aunt Zhang wants to know how many different cubes she can get. Two cubes are considered as the same if and only if one can change to another ONLY by rotating the WHOLE cube. Note that every face of Rubik’s cube is consists of nine small faces. Aunt Zhang can color arbitrary color as she like which means that she doesn’t need to color the nine small faces with same color in a big face. You can assume that Aunt Zhang has 74 different elements to color. (8 points + 12 edges + 9*6=54 small faces)
Input The first line of the date is an integer T, which is the number of the text cases.
Then T cases follow, each case contains one integer K, which is the number of colors. T<=100, K<=100.
Output For each case, you should output the number of different cubes.
Give your answer modulo 10007.
Sample Input 3 1 2 3
Sample Output Case 1: 1 Case 2: 1330 Case 3: 9505
Source
2013 Multi-University Training Contest 4
Recommend zhuyuanchen520
明显是polya计数的题目。
但是本题有74个元素，
总共的变换数数24种。
数起来很麻烦，数据很大。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。。
于是我用代码怒搞之。
写代码找不动点。。。。共24种变换，24种都可以通过多次右旋和上旋得到。。。
300多行代码，，，，，终于找到不动点个数了。。。。。。
附上源代码，。包括找不动点。。
找出来可以直接用公式写的，就没有优化了
1 /* 2 * Author:kuangbin 3 * 1002.cpp 4 */ 5 6 #include <stdio.h> 7 #include <algorithm> 8 #include <string.h> 9 #include <iostream> 10 #include <map> 11 #include <vector> 12 #include <queue> 13 #include <set> 14 #include <string> 15 #include <math.h> 16 using namespace std; 17 const int MOD = 10007; 18 long long pow_m(long long a,long long n) 19 { 20 long long ret = 1; 21 long long tmp = a%MOD; 22 while(n) 23 { 24 if(n&1) 25 { 26 ret *= tmp; 27 ret %=MOD; 28 } 29 tmp *= tmp; 30 tmp %=MOD; 31 n >>= 1; 32 } 33 return ret; 34 } 35 //求ax = 1( mod m) 的x值，就是逆元(0<a<m) 36 long long inv(long long a,long long m) 37 { 38 if(a == 1)return 1; 39 return inv(m%a,m)*(m-m/a)%m; 40 } 41 42 43 int b[100]; 44 int c[100]; 45 46 void rr() 47 { 48 for(int i = 1;i <= 54;i++) 49 c[i] = b[i]; 50 for(int i = 1;i <= 27;i++) 51 b[i+9] = c[i]; 52 for(int i = 1;i <= 9;i++) 53 b[i] = c[27+i]; 54 b[39] = c[37]; 55 b[42] = c[38]; 56 b[45] = c[39]; 57 b[44] = c[42]; 58 b[43] = c[45]; 59 b[40] = c[44]; 60 b[37] = c[43]; 61 b[38] = c[40]; 62 b[41] = c[41]; 63 64 65 b[54] = c[52]; 66 b[51] = c[53]; 67 b[48] = c[54]; 68 b[47] = c[51]; 69 b[46] = c[48]; 70 b[49] = c[47]; 71 b[52] = c[46]; 72 b[53] = c[49]; 73 b[50] = c[50]; 74 75 } 76 void up() 77 { 78 for(int i = 1;i <= 54;i++) 79 c[i] = b[i]; 80 for(int i = 1;i <= 9;i++) 81 b[i] = c[45+i]; 82 for(int i = 37;i <= 45;i++) 83 b[i] = c[i-36]; 84 for(int i = 19;i <= 27;i++) 85 b[i] = c[64-i]; 86 for(int i = 46;i <= 54;i++) 87 b[i] = c[73-i]; 88 89 b[10] = c[12]; 90 b[13] = c[11]; 91 b[16] = c[10]; 92 b[17] = c[13]; 93 b[18] = c[16]; 94 b[15] = c[17]; 95 b[12] = c[18]; 96 b[11] = c[15]; 97 b[14] = c[14]; 98 99 b[28] = c[34]; 100 b[29] = c[31]; 101 b[30] = c[28]; 102 b[33] = c[29]; 103 b[36] = c[30]; 104 b[35] = c[33]; 105 b[34] = c[36]; 106 b[31] = c[35]; 107 b[32] = c[32]; 108 109 } 110 111 bool used[1000]; 112 113 int calc() 114 { 115 memset(used,false,sizeof(used)); 116 117 int ret = 0; 118 for(int i = 1;i <= 54;i++) 119 if(!used[i]) 120 { 121 ret++; 122 int tmp = i; 123 while(!used[tmp]) 124 { 125 used[tmp] = true; 126 tmp = b[tmp]; 127 } 128 } 129 return ret; 130 131 } 132 133 134 void rr2() 135 { 136 for(int i = 1;i <= 12;i++) 137 c[i] = b[i]; 138 b[1] = c[4]; 139 for(int i = 2;i <= 4;i++) 140 b[i] = c[i-1]; 141 b[5] = c[8]; 142 for(int i = 6;i <= 8;i++) 143 b[i] = c[i-1]; 144 b[9] = c[12]; 145 for(int i = 10;i <= 12;i++) 146 b[i] = c[i-1]; 147 } 148 void up2() 149 { 150 for(int i = 1;i <= 12;i++) 151 c[i] = b[i]; 152 b[1] = c[3]; 153 b[2] = c[7]; 154 b[3] = c[11]; 155 b[4] = c[8]; 156 b[5] = c[4]; 157 b[6] = c[2]; 158 b[7] = c[10]; 159 b[8] = c[12]; 160 b[9] = c[1]; 161 b[10] = c[6]; 162 b[11] = c[9]; 163 b[12] = c[5]; 164 } 165 int calc2() 166 { 167 memset(used,false,sizeof(used)); 168 169 int ret = 0; 170 for(int i = 1;i <= 12;i++) 171 if(!used[i]) 172 { 173 ret++; 174 int tmp = i; 175 while(!used[tmp]) 176 { 177 used[tmp] = true; 178 tmp = b[tmp]; 179 } 180 } 181 return ret; 182 183 } 184 185 void rr3() 186 { 187 for(int i = 1;i <= 8;i++) 188 c[i] = b[i]; 189 b[1] = c[4]; 190 for(int i = 2;i <= 4;i++) 191 b[i] = c[i-1]; 192 b[5] = c[8]; 193 for(int i = 6;i <= 8;i++) 194 b[i] = c[i-1]; 195 } 196 void up3() 197 { 198 for(int i = 1;i <= 8;i++) 199 c[i] = b[i]; 200 b[1]=c[4]; 201 b[5]=c[1]; 202 b[8]=c[5]; 203 b[4]=c[8]; 204 b[2]=c[3]; 205 b[3]=c[7]; 206 b[7]=c[6]; 207 b[6]=c[2]; 208 } 209 int calc3() 210 { 211 memset(used,false,sizeof(used)); 212 213 int ret = 0; 214 for(int i = 1;i <= 8;i++) 215 if(!used[i]) 216 { 217 ret++; 218 int tmp = i; 219 while(!used[tmp]) 220 { 221 used[tmp] = true; 222 tmp = b[tmp]; 223 } 224 } 225 return ret; 226 227 } 228 int num1[40]; 229 int num2[40]; 230 int num3[40]; 231 int num[40]; 232 int main() 233 { 234 //freopen("in.txt","r",stdin); 235 //freopen("out.txt","w",stdout); 236 int cnt = 0; 237 for(int i = 0;i < 4;i++) 238 for(int j = 0;j < 4;j++) 239 { 240 for(int x = 1;x <= 54;x++) 241 b[x] = x; 242 for(int x = 0;x < i;x++) 243 up(); 244 for(int x = 0;x < j;x++) 245 rr(); 246 num1[cnt++]=calc(); 247 } 248 for(int x = 1;x <= 54;x++) 249 b[x] = x; 250 rr(); 251 for(int i = 0;i < 4;i++) 252 { 253 num1[cnt++]=calc(); 254 up(); 255 } 256 for(int x = 1;x <= 54;x++) 257 b[x] = x; 258 rr(); 259 rr(); 260 rr(); 261 for(int i = 0;i < 4;i++) 262 { 263 num1[cnt++]=calc(); 264 up(); 265 } 266 cnt = 0; 267 for(int i = 0;i < 4;i++) 268 for(int j = 0;j < 4;j++) 269 { 270 for(int x = 1;x <= 12;x++) 271 b[x] = x; 272 for(int x = 0;x < i;x++) 273 up2(); 274 for(int x = 0;x < j;x++) 275 rr2(); 276 num2[cnt++]=calc2(); 277 //printf("%d\n",calc2()); 278 } 279 280 for(int x = 1;x <= 12;x++) 281 b[x] = x; 282 rr2(); 283 for(int i = 0;i < 4;i++) 284 { 285 num2[cnt++]=calc2(); 286 up2(); 287 } 288 for(int x = 1;x <= 12;x++) 289 b[x] = x; 290 rr2(); 291 rr2(); 292 rr2(); 293 for(int i = 0;i < 4;i++) 294 { 295 num2[cnt++]=calc2(); 296 up2(); 297 } 298 cnt = 0; 299 for(int i = 0;i < 4;i++) 300 for(int j = 0;j < 4;j++) 301 { 302 for(int x = 1;x <= 8;x++) 303 b[x] = x; 304 for(int x = 0;x < i;x++) 305 up3(); 306 for(int x = 0;x < j;x++) 307 rr3(); 308 num3[cnt++]=calc3(); 309 //printf("%d\n",calc3()); 310 } 311 for(int x = 1;x <= 8;x++) 312 b[x] = x; 313 rr3(); 314 for(int i = 0;i < 4;i++) 315 { 316 num3[cnt++]=calc3(); 317 up3(); 318 } 319 for(int x = 1;x <= 8;x++) 320 b[x] = x; 321 rr3(); 322 rr3(); 323 rr3(); 324 for(int i = 0;i < 4;i++) 325 { 326 num3[cnt++]=calc3(); 327 up3(); 328 } 329 for(int i = 0;i < 24;i++) 330 num[i] = num1[i]+num2[i]+num3[i]; 331 //for(int i = 0;i <24;i++) 332 // printf("%d\n",num[i]); 333 int T; 334 scanf("%d",&T); 335 int n ; 336 int iCase = 0; 337 while(T--) 338 { 339 iCase++; 340 scanf("%d",&n); 341 int ans = 0; 342 343 for(int i = 0;i < 24;i++) 344 { 345 ans += pow_m(n,num[i]); 346 ans %=MOD; 347 } 348 ans *= inv(24,MOD); 349 ans %=MOD; 350 printf("Case %d: %d\n",iCase,ans); 351 } 352 return 0; 353 }
