POJ - 1942 D - Paths on a Grid 组合数
来源:互联网 发布:网络代购中心 编辑:程序博客网 时间:2024/05/21 19:40
Paths on a Grid
Time Limit: 1000MS Memory Limit: 30000KTotal Submissions: 25331 Accepted: 6324
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?
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
第一次做这个题是dp做的,学习了组合数,发现组合数更简单
无论怎么走肯定要走n+m这么长的路,在这么长的路里选n个向下走的
#include<stdio.h>#include<string.h>#include<math.h>#include<string>#include<algorithm>#include<queue>#include<vector>#include<map>#include<set>typedef long long ll;using namespace std;ll myc(ll n,ll r){ll sum=1;for(int i=1;i<=r;i++){sum=sum*(n-i+1)/i;}return sum;}int main(){ll a,b;while(~scanf("%lld %lld",&a,&b),a+b){if(a==0||b==0)printf("1\n");elseprintf("%lld\n",myc(a+b,a));}return 0; }
0 0
- POJ - 1942 D - Paths on a Grid 组合数
- POJ 1942 Paths on a Grid 组合数
- poj 1942 Paths on a Grid(组合数模板)
- POJ 1942 Paths on a Grid 组合数的优化
- POJ 1942 Paths on a Grid(组合数公式)
- poj-1942-Paths on a Grid【组合数】
- POJ 1942 Paths on a Grid(求组合数)
- poj 1942 Paths on a Grid (求组合数)
- poj 1942 Paths on a Grid(组合数学--组合数)
- poj1942 Paths on a Grid(组合数)
- poj 1942 Paths on a Grid 求组合数mCn的方法
- 组合数的递归调用:poj 1942 Paths on a Grid
- POJ 1942 Paths on a Grid 组合数的应用 (计算路径总个数)
- poj 1942 Paths on a Grid 组合数学
- Paths on a Grid POJ 1942 组合数学
- POJ 1942 Paths on a Grid(组合数学)
- POJ 1942 Paths on a Grid 组合以及处理阶乘
- [ACM] POJ 1942 Paths on a Grid (组合)
- C++--第五课-继承
- 初学者必看:VC++、Win32 SDK、MFC的区别~
- linux环境下搭建 j2ee环境_0
- centos7 最小安装网络问题解决
- 笔试题10
- POJ - 1942 D - Paths on a Grid 组合数
- cuda《学习笔记三》——共享内存和同步
- nodejs:Express学习笔记(1)
- hdu 1069
- scanf和printf函数
- Codeforces 459C Pashmak and Buses 可重集排列
- 看懂命令行命令的帮助输出
- Java开发的23种涉及模式
- python之斐波那契数列