poj 1942 Paths on a Grid（组合数学--组合数）

Paths on a Grid

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 25146 Accepted: 6274

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

分析：根据自己画出的图，找到规律，这是一个组合数，题目要求C（（n+m），min（n，m））。

因为输入的n和m是无符号整型，数字比较大，所以直接用unsigned long long 了，计算过程中用了double计算中间过程（变量类型很重要！）

算大组合数，可以边乘边除，中间数就不会超出范围，因为题目给出，最后结果保证是无符号整型。有两种变成边除的方法。

对于C（n，m）来举例。因为最后结果为（n*(n-1)*(n-2)....(n-m)）/(m*(m-1)*...*1) 这样可以i = n ，j = m来算，但中间肯定会有浮点数出现，所以要用double。

还有一种是i = n - m ， j = 1 开始算，这样相当于C（n，m） = C（n-1，m-1）* n / m;因为中间每一步都是一个组合数，所以不存在小数的情况，直接可以用ull存储。

代码如下：

第一种方式：

#include <stdio.h>#define ull unsigned long longdouble c(ull x,ull y){double ans = 1.0;ull i,j;for(i=x,j=y;j>=1;j--,i--){ans=ans*i/j;}return ans;}int main(){ull n,m;while(scanf("%I64d %I64d",&n,&m),n||m){ull temp = n;if(temp>m)temp=m;printf("%.0lf\n",c(n+m,temp));}return 0;}

第二种方式：

#include <iostream>using namespace std;unsigned long long c(unsigned long long x,unsigned long long y){    unsigned long long s=1,i,j;    for (i = y + 1, j = 1; i <= x; i++, j++)    {        s = s * i / j;    }    return s;}int main(){    unsigned long long n,m,x;    while(cin>>n>>m)    {        if(n==0&&m==0)        {            break;        }        x=n;        if(x<m)x=m;        cout <<c(n+m,x)<<endl;    }    return 0;}