HDU 1041 Computer Transformation 大数递推

发布时间：2020-12-14 03:30:54 所属栏目：大数据来源：网络整理

导读：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 in

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

2
3

Sample Output

1
1

/*
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1
2).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,sizeof(p));
    memset(dp,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("0n");
    }
    return 0;
}

（编辑：李大同）

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