zoukankan      html  css  js  c++  java
  • POJ 1942, Paths on a Grid

    Time Limit: 1000MS  Memory Limit: 30000K
    Total Submissions: 9478  Accepted: 2376

    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 4
    1 1
    0 0

    Sample Output
    126
    2

     

    Source
    Ulm Local 2002


    // POJ1942.cpp : Defines the entry point for the console application.
    //

    #include 
    <iostream>
    using namespace std;
    double CNM(unsigned int N, unsigned int M)
    {
        M 
    = (M > N - M) ? N - M : M;
        
    double ret = 1;
        
    while(M > 0)ret *= double(N--/ double (M--);
        
    return ret;
    }
    int main(int argc, char* argv[])
    {
        unsigned 
    int A, B;
        
    while(scanf("%d %d"&A, &B) && !(A == 0 && B == 0))
            printf(
    "%.0lf\n",CNM(A + B, B));
        
    return 0;
    }

  • 相关阅读:
    Java设计模式-装饰器模式
    【c++内存分布系列】单独一个类
    【转】LCS
    快速排序
    冒泡排序
    选择排序
    多线程读取全局变量
    【转】一致性hash算法(consistent hashing)
    【转】五笔的字典序编码与解码
    给定一个函数rand()能产生0到n-1之间的等概率随机数,问如何产生0到m-1之间等概率的随机数?
  • 原文地址:https://www.cnblogs.com/asuran/p/1584888.html
Copyright © 2011-2022 走看看