hdu1041【找规律】

Computer Transformation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 7862 Accepted Submission(s): 2942

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：计算机能把1变成01，0变成10，问第n步后00的个数
step0:1
step1:0 1
step2:10 01
step3:0110 1001
step4:1001 0110 0110 1001
step5:0110 1001 1001 0110 1001 0110 0110 1001
第n步的00由n-1步的01变化而来，第n-1步的01则由n-2步的1变化而来
除此之外，第n-2步的00->1010->01100110也贡献一个00
所以F（n）=F（n-2）+2^（n-3）（即n-2步里1的个数）

代码：

#include <iostream>#include <string>#include <cstring>#include <cstdio>#include <cmath>#include <cstdlib>#include <algorithm>#include <queue>#include <map>#define MST(s,q) memset(s,q,sizeof(s))#define INF 0x3f3f3f3f#define MAXN 9999using namespace std;int a[1005][1001];//F(n) = F(n - 2) + 2 ^ (n - 3);void deal(){    MST(a, 0);    a[0][1] = 1;    a[1][1] = 0, a[2][1] = 1, a[3][1] = 1;    for (int i = 4; i <= 1000; i++)    {        int r = 0;        for (int j = 1; j <= 1000; j++)        {            a[0][j] *= 2;            a[0][j] += r;            r = 0;            if (a[0][j] > 9)            {                r = 1;                a[0][j] -= 10;            }        }        r = 0;        for (int j = 1; j <= 1000; j++)        {            a[i][j] = a[i - 2][j] + a[0][j] + r;;            r = 0;            if (a[i][j] > 9)            {                r = 1;                a[i][j] -= 10;            }        }    }}int main(){    deal();    int n;    while (cin >> n)    {        if (n == 1)        {            cout << 0 << endl;            continue;        }        int i = 1000;        while (a[n][i] == 0)i--;        for (; i >= 1; i--)            printf("%d", a[n][i] );        printf("\n");    }}