Asif is a student of East West University and he is currently working for the EWUISP to meet his relatively high tuition fees. One day, as a part of his job, he was instructed to connect cable wires to N houses. All the houses lie in a straight line. He wants to use only the minimum number of cable wires required to complete his task such that all the houses receive the cable service. A house can either get the connection from the main transmission center or it can get it from a house to its immediate left or right provided the latter house is already getting the service.
You are to write a program that determines the number of different combinations of the cable wires that is possible so that every house receives the service.
Example: If there are two houses then 3 combinations are possible as shown in the figure.
Figure: circles represent the transmission center and the small rectangles represent the houses.
Input
Each line of input contains a positive integer N (N ≤ 2000). The meaning of N is described in the above paragraph. A value of 0 for N indicates the end of input which should not be processed.
Output
For each line of input you have to output, on a single line, the number of possible arrangements. You can safely assume that this number will have less than 1000 digits.
Sample Input
1
2
3
0
Sample Output
1
3
8
题链接:UVA10862 Connect the Cable Wires
问题简述:(略)
问题分析:
这个问题的关键是数列的递推式,同时需要注意起始项的值。递推式如下:
f0=1
f1=1
......
fn=3*fn-1 - fn-2。
大数计算是用模拟法来实现的,参见程序。
程序说明:
用Java程序来完成大数计算是最为方便的。
题记:(略)
参考链接:(略)
AC的C++语言程序如下:
AC的C++语言程序(模拟法)如下:
/* UVA10862 Connect the Cable Wires */ #include <bits/stdc++.h> using namespace std; typedef long long LL; const int N = 2000; const int N2 = N / 16; const LL BASE = 1e16; LL fib[N + 1][N2 + 1]; void setfib() { fib[1][0] = 1; fib[2][0] = 3; int len = 1; for(int i = 3; i <= N; i++) { int carry = 0; for(int j = 0; j < len; j++) { fib[i][j] = 3 * fib[i - 1][j] - fib[i - 2][j] + carry; if(fib[i][j] < 0) { carry = -1; fib[i][j] += BASE; } else { carry = fib[i][j] / BASE; fib[i][j] %= BASE; } } if(carry) fib[i][len++] = carry; } } int main() { memset(fib, 0, sizeof(fib)); setfib(); int n; while(~scanf("%d", &n) && n) { int k = N2; while(fib[n][k] == 0) k--; printf("%lld", fib[n][k]); for(int i = k - 1; i >= 0; i--) printf("%016lld", fib[n][i]); printf("\n"); } return 0; }
AC的Java语言程序如下: