| Time Limit: 1000MS | Memory Limit: 30000K | |
| Total Submissions: 25425 | Accepted: 6347 |
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:
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 0Sample Output
1262Source
Ulm Local 2002題目意思：
給出m行n列的矩陣，含m*n個方格，每次可以向上或向右走一步，計算從矩陣左下方走到右上方共有多少種不同的方法。解題思路：
組合數學，向上有m步可走，向右有n步可走，從矩陣左下方走到右上方共要走m+n步，選出m步，則剩下的n步一定是固定的（反之亦然），所以步數是C(m+n,max(m,n))。#include<iostream>#include<cstdio>#include<iomanip>#include<cmath>#include<cstdlib>#include<cstring>#include<map>#include<algorithm>#include<vector>#include<queue>using namespace std;#define INF 0x3f3f3f3f#define MAXN 5050unsigned long long c(unsigned long long x,unsigned long long y)//計算C(x,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(){#ifdef ONLINE_JUDGE#else freopen("F:/cb/read.txt","r",stdin); //freopen("F:/cb/out.txt","w",stdout);#endif ios::sync_with_stdio(false); cin.tie(0); unsigned long long n,m,x; while(cin>>n>>m) { if(n==0&&m==0) break; x=max(m,n); cout<<c(n+m,x)<<endl; } return 0;}/*5 41 10 0*/
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