Given, n, count the number of solutions to the equation x + 2y + 2z = n, where x, y, z, n are non negative integers.
Input
There is at most 1500 inputs. Each input is n (n < 1000001) on a single line.
Output
For each input, output the number of solutions on a single line.
Sample Input
2
3
Sample Output
3
3
问题链接：UVA11296 Counting Solutions to an Integral Equation
问题简述：（略）
问题分析：
x, y, z, n是非负整数，且n < 1000001，x+2y+2z=n。假设x=0，那么2*(y+z)的范围是[0,n]，即y+z的范围是[0,n/2]。y+z确定后x也就确定了。
程序说明：（略）
参考链接：（略）
题记：（略）
AC的C++语言程序如下：
/* UVA11296 Counting Solutions to an Integral Equation */
#include <iostream>
using namespace std;
int main()
{
long long n;
while(~scanf("%lld", &n)) {
n = n / 2 + 1;
printf("%lld\n", n * (n + 1) / 2);
}
return 0;
}