弱校联萌十一大决战之强力热身 E. Rectangle （规律）

E. Rectangle
Time Limit: 1000msMemory Limit: 65536KB 64-bit integer IO format: %lld Java class name: Main
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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

解析：枚举矩形的长h，然后它在h的方向上就有n-h+1种放法，同时宽的最大值w即为k/2 - h，对于每个0~w的宽度wi，它在w方向上的放法有m-wi+1种，求和即为所求方案数

PS：用printf输出long long的结果，你看到的并不是你想要的，竟然是个负数。但是cout还是没问题的

AC代码：

#include 
using namespace std;

int main(){
    #ifdef sxk
        freopen("in.txt", "r", stdin);
    #endif // sxk

    long long n, m, k;
    while(cin>>n>>m>>k){
        k /= 2;
        long long ans = 0;
        for(int h=1; h<=n; h++){
            int w = k - h;
            if(w <= 0) break;
            if(w > m) w = m;
            ans += (n - h + 1) * (m + m-w+1)*w/2;    //对于每个h，在h方向上n-h+1种，在w方向上枚举wi求和为(m + m-w+1) * w / 2种
        }
        cout<