hdoj.1293 The Number of Paths【大数+排列组合】 2015/08/07

The Number of Paths

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 595    Accepted Submission(s): 273

Problem Description

Let f (n) be the number of paths with n steps starting from O (0, 0), with steps of the type (1, 0), or (-1, 0), or (0, 1), and never intersecting themselves. For instance, f (2) =7, as shown in Fig.1. Equivalently, letting E=(1,0),W=(-1,0),N=(0,1), we want the number of words A1A2...An, each Ai either E, W, or N, such that EW and WE never appear as factors.

 

Input

There are multiple cases in this problem and ended by the EOF. In each case, there is only one integer n means the number of steps(1<=n<=1000).

 

Output

For each test case, there is only one integer means the number of paths.

 

Sample Input

12

 

Sample Output

37

 

Author

SmallBeer (CML)

 

Source

杭电ACM集训队训练赛（VIII）  

注：f(n) = 2*f(n-1)+f(n-2)

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int p[1010][500];int main(){    //int p[1010][500];    memset(p,0,sizeof(p));    p[0][1] = 1;    p[1][1] = 3;    p[0][0] = p[1][0] = 1;    for( int i = 2 ; i <= 1000 ; ++i ){        for( int j = 1 ; j <= p[i-1][0] ; ++j )            p[i][j] = p[i-1][j] * 2 + p[i-2][j];        for( int j = 1 ; j <= p[i-1][0] ; ++j ){            if( p[i][j] > 9 ){                p[i][j+1] += p[i][j] / 10;                p[i][j] %= 10;            }        }        p[i][0] = p[i][p[i-1][0]+1] ? p[i-1][0] + 1 : p[i-1][0] ;    }    int n;    while(cin>>n){        for( int i = p[n][0] ; i > 0 ; --i )            printf("%d",p[n][i]);        printf("\n");    }    return 0;}