1023

 

  1 /*
  2 Problem Description
  3 As we all know the Train Problem I, the boss of the Ignatius Train Station want to know if all the trains come in strict-increasing order, how many orders that all the trains can get out of the railway.
  4 
  5  
  6 
  7 Input
  8 The input contains several test cases. Each test cases consists of a number N(1<=N<=100). The input is terminated by the end of file.
  9 
 10  
 11 
 12 Output
 13 For each test case, you should output how many ways that all the trains can get out of the railway.
 14 
 15  
 16 
 17 Sample Input
 18 1
 19 2
 20 3
 21 10
 22  
 23 
 24 Sample Output
 25 1
 26 2
 27 5
 28 16796
 29 
 30 HintHint 
 31 The result will be very large, so you may not process it by 32-bit integers.
 32  
 33 
 34 Author
 35 Ignatius.L
 36 
 37  */
 38  /*Description: to 
 39  * 
 40  * Author: Vincent
 41  * 
 42  * Date:2010/04/
 43  * Contact:agilely@126.com 
 44  * Zju CS Lab
 45  */
 46 //按1、2、3，，，来的车，给出车数，求最大出站种数
 47 #include <stdio.h>//查表。。结果事先算完放入sum[]
 48 #include <string.h>
 49 char temp[1100]={0},sum[1100]={0};
 50 int main()
 51 {
 52     void multipe(char a[],char b[]);
 53     void puls(char a[],char b[]);//函数声明
 54     char result[110][1000] = {0};
 55     char tt[1100],hh[1100];
 56     int i,j,k;
 57     int n;
 58     result[0][0] = 49;//49为ASCII的1
 59     result[1][0] = 49;
 60     for( i=2; i <= 101; i ++)
 61     {
 62         result[i][0]='0';
 63         for(j = 0;j < i;j ++)
 64         {
 65             memset(tt,0,sizeof(tt));
 66             memset(hh,0,sizeof(hh));
 67             strcpy(tt,result[j]);
 68             strcpy(hh,result[i-j-1]);
 69             multipe(tt,hh);
 70             puls(temp,result[i]);
 71             memset(result[i],0,sizeof(result[i]));
 72             for(k=0;sum[k];k++)
 73                 result[i][k]=sum[k];
 74         }
 75     }
 76     while(scanf("%d",&n)!=EOF && n != -1)
 77         printf("%s\n",result[n]);
 78     return 0;
 79 }
 80 void multipe(char a[],char b[])
 81 {
 82     int la,lb;
 83     char t[1000];
 84     int c[2000];
 85     int i,j,l;
 86     la=strlen(a);
 87     lb=strlen(b);
 88     memset(t,0,sizeof(t));
 89     strcpy(t,a);
 90     for(j=0,i=la-1;i>=0;i--) a[j++]=t[i];
 91     memset(t,0,sizeof(t));
 92     strcpy(t,b);
 93     for(j=0,i=lb-1;i>=0;i--) b[j++]=t[i];
 94     memset(c,0,sizeof(c));
 95     for(i=0;i<la;i++)
 96         for(j=0;j<lb;j++)
 97         {
 98             c[i+j]+=(a[i]-48)*(b[j]-48);
 99             c[i+j+1]+=c[i+j]/10;
100             c[i+j]%=10;
101         }
102     l=la+lb+1;
103     while(l>=0&&!c[l]) l--;
104     if(l<0)    c[0]=0;
105     memset(temp,0,sizeof(temp));
106     if(l<0)    temp[0]='0';
107     for(i=l,j=0;i>=0;i--,j++)
108         temp[j]=c[i]+48;
109 }
110 void puls(char a[],char b[])
111 {
112     int i,j,k1,la,lb,s,lsum;
113     int c=0;
114     char h;
115     for(j=0;a[j];j++); la=j-1;
116     for(j=0;b[j];j++); lb=j-1;
117     memset(sum,0,sizeof(sum));
118     k1=0;
119     while(1){
120         if(la<0&&lb<0) break;
121         else if(la<0) s=b[lb--]-48;
122         else if(lb<0) s=a[la--]-48;
123         else s=a[la--]-48+b[lb--]-48;
124 
125         if(k1==0) {sum[k1]=s%10+48 ;c=s/10;}
126         else {sum[k1]=(s+c)%10+48;c=(s+c)/10;}
127         k1++;
128     }
129     if(c!=0) sum[k1]=c+48;
130     lsum=strlen(sum)-1;
131     for(i=0;i<=lsum/2;i++)
132         {h=sum[i];sum[i]=sum[lsum-i];sum[lsum-i]=h;}
133 }
134 
135

 

posted @ 2010-04-15 23:22  にんじゃ  阅读(266)  评论(0)    收藏  举报