Computer Transformation(简单数学题+大数)

H - Computer Transformation

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice _

Appoint description: 
System Crawler  (Oct 10, 2016 1:02:59 PM)

Description

A sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simultaneously tranforms each digit 0 into the sequence 1 0 and each digit 1 into the sequence 0 1. So, after the first time step, the sequence 0 1 is obtained; after the second, the sequence 1 0 0 1, after the third, the sequence 0 1 1 0 1 0 0 1 and so on.

How many pairs of consequitive zeroes will appear in the sequence after n steps?

Input

Every input line contains one natural number n (0 < n ≤1000).

Output

For each input n print the number of consecutive zeroes pairs that will appear in the sequence after n steps.

Sample Input

2
3

Sample Output

1

1

//这题意思是

1 01        0

2 1001       1

3 01101001    1

1->01 0->10

就这么一直变下去,问 n 步之后，有多少个相邻的 0

我是先输出模拟了大概 8 项找出了规律，然后。。。wa

因为数据太大，long long 也存不下

然后用字符串，当做大数处理，就可以了，注意一些细节问题，代码里有

 /*
 //EEE
 #include <stdio.h>
 #include <string.h>
 
 char ch[10][1200];
 
 int main()
 {
     strcpy(ch[0],"01");
 
     for (int i=1;i<10;i++)
     {
         int k=0;
         int len=strlen(ch[i-1]);
         for (int j=0;j<len;j++)
         {
             if (ch[i-1][j]=='0')
             {
                 ch[i][k++]='1';
                 ch[i][k++]='0';
             }
             else
             {
                 ch[i][k++]='0';
                 ch[i][k++]='1';
             }
         }
         ch[i][k]='\0';
     }
     for (int i=0;i<10;i++)
     {
 
         int sum=0;
         int len =strlen(ch[i]);
         for (int j=0;j<len-1;j++)
         {
             if (ch[i][j]=='0'&&ch[i][j+1]=='0')
                 sum++;
         }
         //printf("%s\n",ch[i]);
         printf("%d\n",sum);
     }
     return 0;
 }
 */
 
 #include <stdio.h>
 #include <string.h>
 #include <math.h>
 #include <algorithm>
 using namespace std;
 
 char num[][];
 char str[];
 
 char *add(char s[])
 {
     memset(str,,sizeof(str));
 
     int i;
     int len=strlen(s);
     for (i=;i<len;i++)
         s[i]-='';
     reverse(s,s+len);
 
     for (i=;i<len;i++)
     {
         str[i]+=s[i]*;
         if (i==) str[i]++;
 
         if (str[i]>=)
         {
             str[i]-=;
             str[i+]++;
         }
 
     }
     while (str[i]!=) i++;
     str[i]='\0';
     for (int j=;j<i;j++) str[j]+='';
     for (int j=;j<len;j++) s[j]+='';
     reverse(str,str+i);
     reverse(s,s+len);
     return str;
 }
 
 char *de(char s[])
 {
     memset(str,,sizeof(str));
 
     int i;
     int len=strlen(s);
     for (i=;i<len;i++)
         s[i]-='';
     reverse(s,s+len);
 
     int k=,flag=;
     for (i=;i<len;i++)
     {
         str[i]+=s[i]*;
         if (str[i]>=)
         {
             str[i]-=;
             str[i+]++;
         }
 
         if (flag==&&str[i]==)//减1，看这个数最后有几个 0
         {
             k++;
         }
         if (str[i]!=) flag=;
 
     }
     while (str[i]!=) i++;
     str[i]='\0';
 
     str[k]--;
     while (k--) str[k]=;//减 1
 
     for (int j=;j<i;j++) str[j]+='';
     for (int j=;j<len;j++) s[j]+='';
     reverse(str,str+i);
     reverse(s,s+len);
     return str;
 }
 
 int main()
 {
     int n;
     int i;
     int xx=;
 
     strcpy(num[],"");
 
     for (i=;i<=;i++)
     {
         if (i%==)
         {
             add(num[i-]);
             strcpy(num[i],str);
         }
 
         else
         {
             de(num[i-]);
             strcpy(num[i],str);
         }
     }
     while (scanf("%d",&n)!=EOF)
     {
         printf("%s\n",num[n]);
     }
     return ;
 }