poj 1942(裸组合数+隐式类型转换）

Paths on a Grid

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 20134 Accepted: 4890

Description

Imagine you are attending your math lesson at school. Once again, you are bored because your teacher tells things that you already mastered years ago (this time he's explaining that (a+b)²=a²+2ab+b²). So you decide to waste your time with drawing modern art instead. 

Fortunately you have a piece of squared paper and you choose a rectangle of size n*m on the paper. Let's call this rectangle together with the lines it contains a grid. Starting at the lower left corner of the grid, you move your pencil to the upper right corner, taking care that it stays on the lines and moves only to the right or up. The result is shown on the left: 

Really a masterpiece, isn't it? Repeating the procedure one more time, you arrive with the picture shown on the right. Now you wonder: how many different works of art can you produce?

Input

The input contains several testcases. Each is specified by two unsigned 32-bit integers n and m, denoting the size of the rectangle. As you can observe, the number of lines of the corresponding grid is one more in each dimension. Input is terminated by n=m=0.

Output

For each test case output on a line the number of different art works that can be generated using the procedure described above. That is, how many paths are there on a grid where each step of the path consists of moving one unit to the right or one unit up? You may safely assume that this number fits into a 32-bit unsigned integer.

Sample Input

5 41 10 0

Sample Output

1262

Source

Ulm Local 2002

刚开始写了一个dp来搞，必然直接爆内存，因为如果当m或者n里面有一个为1的时候，内存将达到2^32

其实就直接是C(m+n, min(m, n)) 这里要算一个min，否则也可能会超时。因为如果你试图去计算C(2^32, 2^32-1)的时候，必然会超时。

注意可以用unsigned int来读取数据，但是计算组合数的函数中，一定要用double，否则可能会损失精度。C语言会做隐式类型转换。

中间wrong了一次是因为输入数据的类型定义为了int，导致溢出。

提交记录：

1、MLE 因为用dp当m或n比较小时，可能会超内存。

2、WA 因为输入搞成了int，溢出了。

3、AC

代码（注意代码中的隐式转换是work的）

这里有三个技巧注意一下：

计算组合数其实在循环内一行语句就可以搞定。

一个浮点数四舍五入其实就是先加0.5再强制转化。

判断m和n同时为0则break：if (!m && !n) break;