联萌十一大决战之强力热身 E. Rectangle

[题目链接] (http://www.bnuoj.com/v3/contest_show.php?cid=6865#problem/E)
题目描述：
E. Rectangle
Time Limit: 1000msMemory Limit: 65536KB 64-bit integer IO format: %lld Java class name: Main
Submit Status
frog has a piece of paper divided into n rows and m columns. Today, she would like to draw a rectangle whose perimeter is not greater than k.

There are 8 (out of 9) ways when n=m=2,k=6

Find the number of ways of drawing.
Input
The input consists of multiple tests. For each test:

The first line contains 3 integer n,m,k (1≤n,m≤5⋅104,0≤k≤109).
Output
For each test, write 1 integer which denotes the number of ways of drawing.
Sample Input
2 2 6
1 1 0
50000 50000 1000000000
Sample Output
8
0
1562562500625000000
言简意赅：
在n*m的矩阵中画蓝图，要保证使得蓝图为矩形，且该矩形的边长小于等于K，此题也算得上是水题了，整场比赛下来也是正确率偏高的一个题了，但东施般的正常暴力方法是完全不可能过的，因为数据颇大，所以此题应采用暴力一条边的方法，在暴力一条边的同时，另一条边的情况会相应的体现出来，很有技巧的暴力，别样的思维方式，值得学习。
代码实现如下：

#include<iostream>
#define LL long long
using namespace std;
int main()
{
    LL n,m,k;
    while(cin >> n >> m >> k)
    {
        LL ans = 0;
        for(int i=1; i<=n && k/2-i>0; i++)
        {
            LL a,b;
            if (k/2-i<m)
            {
                a = n-i+1;
                b = (m+(m-k/2+i+1))*(k/2-i)/2;
                ans+=a*b;
            }
            else
            {
                a = n-i+1;
                b = (m+1)*m/2;
                ans+=a*b;
            }
        }
        cout<<ans<<endl;
    }
    return 0;
}