Rectangle

frog has a piece of paper divided into nn rows and mm columns. Today, she would like to draw a rectangle whose perimeter is not greater than kk .

There are 88 (out of 99 ) ways when n=m=2,k=6n=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 33 integer n,m,kn,m,k (1≤n,m≤5?104,0≤k≤1091≤n,m≤5?104,0≤k≤109 ).

Output

For each test, write 11 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

这题我看到一个题解，感觉写的很透彻，放这存一下。

题意：给定长度，求在不大于这个长度下，有多少个矩形(矩形周长不大于给定长度)。主要是用到了矩形的对称性  以及以下这个性质  在长为n，宽为m的矩形上，长为i，宽为j的矩阵个数为(n-i+1)x(m-j+1)。

证明：首先考虑n在一个长为n的矩形中从1~i,2~i+1,3~i+2,n-i+1~n;分别为长为i的矩形同理考虑m宽为j的矩形1~j,2~j+1,3~j+2,m-j+1~m;这样的话在1~j下就有n-i+1个矩形所以总共就是(n-i+1)x(m-j+1)；

那么这道题的答案就出来了  记num=k/2-i (num>0)  k为周长  i为长  num为宽  在num<=m时  num可以取1,2,3,…,num  所以答案为  ans=(n-i+1)x(m-1+1)+(n-i+1)x(m-2+1)+…+(n-i+1)*(m-num+1);  提取(n-i+1),就是一个等差数列  所以  ans+=(n-i+1)x(2m-num+1)num/2;  当num>m时  num替换为m  ans+=(n-i+1)x(2m-m+1)m/2;  ans+=(n-i+1)x(m+1)m/2;  代码如下

#include<cstdio>
#define ll long long
ll n,m,k,ans,num;
int main()
{
  while(~scanf("%lld%lld%lld",&n,&m,&k))
  {
    ans=0;
    for(int i=1;i<=n;i++)
    {
      num=k/2-i;
      if(num<=m&&num>0)ans+=(n-i+1)*(2*m-num+1)*num/2;
      else if(num>0)ans+=(n-i+1)*(1+m)*m/2;
    }
    printf("%lld\n",ans);
  }
  return 0;
}

题解地址：http://blog.csdn.net/VictorZC8/article/details/51242491