Paths on a Grid

time limit per test 1 second memory limit per test 256 megabytes
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=a 2+2ab+b 2). 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?

传送门:POJ1942

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

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5 4
1 1
0 0

Sample Output

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126
2

题解

传统的dp在这行不通
这里就用到了组合数学的公式
结果是C(n,n+m) 因为每一步都只能向右或者向上 而且向上的步数等于行数 那么就是n+m个地方选n个地方放向上,剩下的自然就是向右的了
然后组合公式注意每项上下除最后取整加速

AC code:(不包含输入类)

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import java.io.*;
import java.util.*;
public class Main {

public static void main(String[] args) {
FastScanner sc=new FastScanner();
PrintWriter pw=new PrintWriter(System.out);
while(sc.hasNext()){
long n=sc.nextLong();
long m=sc.nextLong();
if(n==0&&m==0)break;
//C(n,m+n)
pw.println(C(n,n+m));
pw.flush();
}

}
static long C(long a,long b){
//b!/(a!*(b-a)!)
long t=Math.min(a,b-a);
double result=1.0;
while(t>0){
result*=(b*1.0/t);
b--;
t--;
}
return (long)(result+0.5);
}
}