HDU 1041.Computer Transformation【递推+高精】【12月27】

Computer Transformation

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

23

 

Sample Output

11

直接上代码：

#include<iostream>#include<cstdio>#include<algorithm>#include<cstring>#include<string>#include<map>using namespace std;int main(){    long long  f[1010][100]={0}, a[1010];    //n为奇数 f[n] = f[n-1]*2 + 1;    //n为偶数 f[n] = f[n-1]*2 -1;    int n;    bool flag = true;    f[1][0] = 0;    a[1] = 1;    for(int i = 2;i < 1010; ++i)    {        a[i] = a[i-1];        for(int j = 0;j < a[i]; ++j)    f[i][j] = f[i-1][j]*2;        for(int j = 0;j < a[i]; ++j)        {            f[i][j+1] += f[i][j]/10000;            f[i][j] %= 10000;        }        if(flag) f[i][0] +=1;        else f[i][0] -= 1;        int k = 0;        while(f[i][k] < 0)        {            f[i][k] += 10000;            f[i][k+1] -= 1;            k++;        }        if(f[i][a[i]] > 0) a[i]++;        flag = !flag;    }    while(scanf("%d", &n) != EOF)    {        cout <<f[n][a[n]-1];        for(int i = a[n]-2;i >= 0; --i)            printf("%04d",f[n][i]);        cout << endl;    }    return 0;}