计算建造建筑物的可能方法
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
输入输出
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。
算法
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示例
#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。