Computer Transformation（规律，大数打表）

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6946    Accepted Submission(s): 2515

Problem 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

题解：f(n)=2*f(n-2)+f(n-1)由于每个n-2的00生成两个01，在n的地方就会生成两个00；n-1的00在n生成一个00；由此可以找出规律；

代码：

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;
 #define mem(x,y) memset(x,y,sizeof(x))
 typedef long long LL;
 const int MAXN=;
 char dp[MAXN][MAXN];
 char c[MAXN];
 int x[MAXN],y[MAXN],z[MAXN];
 void bigsum(char *a,char *b){
     int len1,len2;
     mem(x,);mem(y,);mem(z,);
     len1=strlen(a);len2=strlen(b);
     int len=max(len1,len2);
     for(int i=len1-,j=;i>=;i--,j++)x[j]=a[i]-'';
     for(int i=len2-,j=;i>=;i--,j++)y[j]=b[i]-'';
     for(int i=;i<len;i++){
         z[i]=x[i]+y[i]+z[i];
         z[i+]+=z[i]/;
         z[i]%=;
         if(z[len])len++;
     }
     for(int i=len-,j=;i>=;i--,j++)c[j]=z[i]+'';
     c[len]='\0';
 }
 int main(){
     int n;
     dp[][]='';dp[][]='\0';
     dp[][]='';dp[][]='\0';
     for(int i=;i<=;i++){
         bigsum(dp[i-],dp[i-]);
         bigsum(c,dp[i-]);
         memcpy(dp[i],c,sizeof(c));
     }
     while(~scanf("%d",&n)){
         printf("%s\n",dp[n]);
     }
     return ;
 }