ACM: SCU 4440 Rectangle

 SCU 4440 Rectangle

Time Limit:0MS     Memory Limit:0KB     64bit IO Format:%lld & %llu

 Practice

Description

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
　　
　　一开始以为是dp，但是数据会爆掉，然后放在一边了。。
　　后来其他题目不会写了后，回过来想这题，想了好久是暴力还是数学计算。。
　　虽然最后用的是暴力加数学计算。。。
　　单纯暴力坑定会超时，仔细想了下，可以暴力去枚举一条边，再去考虑另一条的具体情况。
    最后在模拟另一条边的情况的时候推算要注意一点。
　　
　　AC代码：

#include"algorithm"
#include"iostream"
#include"cstring"
#include"cstdlib"
#include"cstdio"
#include"string"
#include"vector"
#include"queue"
#include"cmath"
#include"map"
using namespace std;
typedef long long LL ;
#define memset(x,y) memset(x,y,sizeof(x))
#define memcpy(x,y) memcpy(x,y,sizeof(x))
#define FK(x) cout<<"["<<x<<"]
"
#define bigfor(T)  for(int qq=1;qq<= T ;qq++)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

int main() {
	LL n,m,k;
	while(cin>>n>>m>>k) {
		k>>=1;
		LL ans = 0;
		for(int i=1; i<=n&&k-i>0; i++) {
			LL x=min(k-i,m);
			LL a=n-i+1;
			LL b=(m+m-x+1)*x/2;
			ans+=a*b;
		}
		cout<<ans<<endl;
	}
	return 0;
}