这里给出n个路段,每个路段在建造建筑物的过程中有两个侧面。如果需要在两栋房屋之间留出一个空的空间,那么有多少种方法可以在地块中建造建筑物。
建造建筑物有四种可能性
路的一侧
路的另一边
不能建造建筑物
道路两边
Input: It takes the number of sections to construct buildings. Say the input is 3. Output: Enter Number of sections: 3 Buildings can be constructed in 25 不同的方法。
constructionWays(n)
输入:有n个部分。
输出-可能的方法数量。
Begin if n = 1, then return 4 countEnd := 1 countEndSpace := 1 for i := 2 to n, do prevCountEnd := countEnd prevCountEndSpace := countEndSpace countEndSpace := countEnd + prevCountEndSpace countEnd := prevCountEndSpace done answer := countEndSpace + countEnd return answer^2 End
#include<iostream> using namespace std; int constructionWays(int n) { if (n == 1) //if there is one section return 4; //4 possible ways to construct building in that section //设置末端的计数值,并以空格结尾 int countEnd=1, countEndSpace=1, prevCountEnd, prevCountEndSpace; for (int i=2; i<=n; i++) { //fot the second section to nth section prevCountEnd = countEnd; prevCountEndSpace = countEndSpace; countEndSpace = countEnd + prevCountEndSpace; countEnd = prevCountEndSpace; } //以空间和建筑结尾的可能方法 int answer = countEndSpace + countEnd; return (answer*answer); //for two sides the answer will be squared } int main() { int n; cout << "Enter Number of sections: "; cin >> n; cout << "Buildings can be constructed in " << constructionWays(n) <<" 不同的方法。" ; }
输出结果
Enter Number of sections: 3 Buildings can be constructed in 25 不同的方法。