典型的高精度题，本着“和平共处五项原则”，我抄了一下ZXPN的代码，不过略加改进，中间也出了点问题，忘了算fac[0]了。

1 #include
     
        2 #include
      
         3 char dp[105][105][500],fac[105][200];  4 void add(char *s1,char *s2,char *s3)  5 {
   //s3 = s1 + s2  6     int len1 = strlen(s1);  7     int len2 = strlen(s2);  8     int a[10000];  9     int k = len1 > len2 ? len1 : len2; 10     int i,j; 11     for(i = 0;i <= k;i++) 12         a[i] = 0; 13     for(i = len1 - 1,j = k;i >= 0;i--,j--) 14     { 15         a[j] = s1[i] - '0'; 16     } 17     for(i = len2 - 1,j = k;i >= 0;i--,j--) 18     { 19         a[j] = a[j] + s2[i] - '0'; 20         a[j - 1] = a[j - 1] + a[j] / 10; 21         a[j] %= 10; 22     } 23     if(a[j] >= 10) 24     { 25         a[j - 1] = a[j] / 10; 26         a[j] %= 10; 27     } 28     i = 0; 29     if(a[0] == 0) 30         i = 1; 31     for(j = i;j <= k;j++) 32         s3[j - i] = a[j] + '0'; 33     s3[j - i] = '\0'; 34 } 35 void mul(char *s1,char *s2,char *s3) 36 {
   //s3 = s1 * s2 37     int len1 = strlen(s1); 38     int len2 = strlen(s2); 39     int a[10000]; 40     int k = len1 + len2 - 1; 41     int i,j; 42     for(i = 0;i <= k;i++) 43         a[i] = 0; 44     for(i = len1 - 1;i >= 0;i--) 45     { 46         for(j = len2 - 1;j >= 0;j--) 47         { 48             a[i + j + 1] += (s1[i] - '0') * (s2[j] - '0'); 49             a[i + j] += a[i + j + 1] / 10; 50             a[i + j + 1] %= 10; 51         } 52     } 53     if(a[1] >= 10) 54     { 55         a[0] = a[1] / 10; 56         a[1] %= 10; 57     } 58     i = 0; 59     if(a[0] == 0) 60         i = 1; 61     for(j = i;j <= k;j++) 62         s3[j - i] = a[j] + '0'; 63     s3[j- i] = '\0'; 64 } 65 void facmul(char *s1,int n,char *s2) 66 {
   //s2 = s1 * n 67     int l,k,i,j,t,f=0; 68     l = strlen(s1); 69     k = l + 2;//本题n最大为100 70     for(i = 0; i < k; i++) 71         s2[i] = 0; 72     for(i = l-1,j = k-1; i >= 0; i--,j--) 73     { 74         t = (s1[i]-'0') * n; 75         s2[j] += t % 10; 76         s2[j-1] += t / 10; 77         if(s2[j] > 9) 78             s2[j-1] += s2[j]/10,s2[j] %= 10; 79     } 80     if(s2[1] > 9) 81         s2[0] += s2[1]/10,s2[1] %= 10; 82     while(s2[f] == 0) 83         f++; 84     for(i = 0; i < k-f; i++) 85         s2[i] = s2[i+f]+'0'; 86     s2[k-f] = '\0'; 87 } 88 int main() 89 { 90     int i,j,a,b,n=1; 91     strcpy(fac[0],"1"); 92     for(i = 1; i <= 100; i++) 93         facmul(fac[i-1],i,fac[i]); 94     for(i=0;i<=100;i++) 95         strcpy(dp[i][0],"1"); 96     for(i=1;i<=100;i++) 97         for(j=1;j<=i;j++) 98             if(i==j) 99                 strcpy(dp[i][j],dp[i][j-1]);100             else101                 add(dp[i][j-1],dp[i-1][j],dp[i][j]);102     for(i = 0;i <= 100;i++)103         for(j = 0;j <= i;j++)104         {105             mul(dp[i][j],fac[i],dp[i][j]);106             mul(dp[i][j],fac[j],dp[i][j]);107         }108     while(~scanf("%d%d",&a,&b),a||b)109     {110         printf("Test #%d:\n",n++);111         if(a < b)112             printf("0\n");113         else114             printf("%s\n",dp[a][b]);115     }116     return 0;117 }