Description
In mathematics, a polygonal number is a number represented as dots or pebbles arranged in the shape of a regular polygon. The dots are thought of as alphas (units). These are one type of 2-dimensional figurate numbers. The following picture shows how triangular numbers, square numbers, pentagonal numbers and hexagonal numbers represented as dots arranged in the shape of corresponding regular polygon.
2016 is not only a leap year but also a triangular and hexagonal year. If you are patient enough, you can count the number of the dots in the left triangle or in the right hexagon in the following picture. The number of dots in each shape is 2016.
Therefore, 2016 is a triangular-hexagonal-leap year. The previous triangular-hexagonal-leap year is 1540 and the next is 2556. So living to see 2016 is very rare experience.
You task is to list the triangular-hexagonal-leap years from 2016 to 990528. 990528 is also a triangular-hexagonal-leap year.
Input
This problem has no input.
Output
Please print each triangular-hexagonal-leap year in increasing order.
For example, if you are asked to list the triangular-hexagonal-leap years from 780 to 2556, the output should be:
780 1128 1540 2016 2556
Sample Output
2016
2556
... <-- some lines are skipped
990528
题意：给出三四五六边型数定义，然后让你输出指定范围的三六边型和闰年数；
解题思路：这题主要是时间限制问题和怎么判断是否是三六边型数难办，开始的时候李远飞大师（忍住不笑）暴力打表但是神之WA了，后来我把方程解出来，比如判断三角形数：令x=[n*(n+1)]/2;解出来n=(1+sqrt(1+8*i))/2.0;只需要判断n是否为整数就可以了；上学期第一套提有一个类似解方程的题，哎~天才的记忆力，嘻嘻嘻^.^;
感悟：长时间的沉淀，一瞬间的灵光一闪~~~
代码：
#include
#include
#include
using namespace std;
bool leap(int i)
{
if((i%4==0&&i0!=0)||(i%4==0&&i@0==0))
return true;
return false;
}
bool check(int i)
{
double a,b;
a=(1+sqrt(1+8*i))/2.0;
b=(1+sqrt(1+8*i))/4.0;
if(a-(int) a<1e-6&&b-(int) b<1e-6&&leap(i))//是不是整数
return true ;
return false;
}
int main()
{
for(int i=2016;i<=990528;i++)
if(check(i))
printf("%d
",i);
}