POJ 1942 Paths on a Grid【组合数学】

Paths on a Grid

Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 24879 Accepted: 6185

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

感觉数学废掉了。。。。。

对于一个n*m的网格，一定是n次向上走，m次向下走，只是向下和向上的顺序的问题。也就是走n+m个空的位置，让你用n个向上，m个向上去填满，求解共有几种方法去填满。然后用组合数学处理。

#include <map>#include <set>#include <cstdio>#include <cstring>#include <algorithm>#include <queue>#include <iostream>#include <stack>#include <cmath>#include <string>#include <vector>#include <cstdlib>//#include <bits/stdc++.h>//#define LOACL#define space " "using namespace std;typedef long long LL;//typedef __int64 Int;typedef pair<int, int> paii;const int INF = 0x3f3f3f3f;const double ESP = 1e-5;const double PI = acos(-1.0);const int MOD = 1e9 + 7;const int MAXN = 1e5 + 10;LL myc(LL x,LL y) {LL ans = 1, i;    for (i = 1; i <= y; i++) {        ans = ans*(x + 1 - i)/i;    }    return ans;}int main() {LL n, m;while (scanf("%lld%lld", &n, &m) != EOF) {if (n == 0 && m == 0) break; //坑1: WALL minn = min(n, m); //坑2：TLEprintf("%lld\n", myc(n + m, minn));}return 0;}