JOJ 1007解题
1007: Triangles
Result | TIME Limit | MEMORY Limit | Run Times | AC Times | JUDGE |
---|---|---|---|---|---|
3s | 8192K | 7020 | 2611 | Standard |
A triangle can be made up out of dots, like the ones shown below:
The number of dots in each one is called a triangular number. More precisely, a number is said to be triangular if it is of the form ½ n(n+1). The numbers 1, 3, 6, 10, and 15 are all triangular.
Given a number, determine if it is triangular. If it is, draw the corresponding right triangle, if it is not, say so.
Input
A positive integer less than 2000. Each case will be on a separate line. A zero (0) denotes the end of input.
Output
The phrase “<num> is a triangular number.” followed by a right triangle of “*” characters, or the phrase “<num> is not a triangular number.” The triangle should be oriented with the hypotenuse on the right and the right angle in the lower left.
Sample Input
3 4 6 0
Sample Output
3 is a triangular number. * ** 4 is not a triangular number. 6 is a triangular number. * ** ***
代码如下:
#include <iostream> #include <queue> #include <cmath> using namespace std; bool isTriangle(int n) { n = 2*n; int srt = (int)sqrt(n); if (srt * (srt + 1) == n) return true; return false; } void printTriangle(int n) { for (int i = 1 ; i <= n; i ++) { for(int j = 1; j <= i ; j ++) cout<<"*"; cout<<endl; } } int main() { queue<int> input; int a; while (cin>>a && a!=0) { input.push(a); } while (!input.empty()) { if(isTriangle(input.front())) { cout << input.front() << " is a triangular number."<<endl; printTriangle((int)sqrt(input.front()*2)); } else { cout << input.front() << " is not a triangular number."<<endl; } input.pop(); } return 0; }主要解题思路:
如果一个数是三角数,那么一定有N= 1/2*n(n+1),因此就有:2*N = n*(n + 1);
可以推出:n < sqrt(n*(n+1)) < n + 1,所以我们可以得出:n = (int)sqrt(n*(n+1))(注:截取整数部分,这个结果一定不是一个整数)
然后再验证是否有:n*(n + 1) == 2*N,如果是,那么N就是三角数,否则就不是三角数。
知道了这些就不难理解上面的解题思路了!!
有不妥之请留言,共同探讨,共同进步!