HDU 1041 Computer Transformation 大数递推

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5787    Accepted Submission(s): 2090

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

/*要学会分析题目啊。。。。。00的产生必然由n-1组的01产生，即00的个数等于n-1组01的个数。n-1组01的产生有两个来源。1).由n-2组的12).由n-2组的00变幻1010，其中一个00只能产生一个01.这个就是要求的n-2连续的个数 可以推出第n组00的个数dp[n]=dp[n-2]+第n-2组1的个数(2^((n-2)-1)); 发现还要用大数 */#include<iostream>#include<cstring>using namespace std;int dp[1002][500];//因为两个长度 比较麻烦 直接500 不记长度 int p[500];int main(){        int n,i,j,z,t;        memset(p,0,sizeof(p));    memset(dp,0,sizeof(dp));    dp[1][1]=0;    dp[2][1]=1;    dp[3][1]=1;        p[1]=1;     for(i=4;i<1001;i++)    {        for(j=1;j<500;j++)//处理2的多次方                p[j]=p[j]*2;        z=0;        for(j=1;j<500;j++)            {            t=p[j]+z;            z=t/10;            p[j]=t%10;        }                    for(j=1;j<500;j++)//处理递推方程             dp[i][j]=dp[i-2][j]+p[j];                    z=0;        for(j=1;j<500;j++)            {            t=dp[i][j]+z;            z=t/10;            dp[i][j]=t%10;        }    }    while(cin>>n)    {        i=499;        while(dp[n][i]==0) i--;        for(;i>=1;i--)            printf("%d",dp[n][i]);        if(n!=1)            printf("\n");        else             printf("0\n");    }    return 0;}