poj1942(Paths on a Grid)

题目地址:Paths on a Grid

题目大意:

    给你一个矩形的格子,让你从左下角走到右上角,每次移动只能向上或者向右,问你有多少种可能的路径。

解题思路:

     水题,排列组合。推出公式C(m+n,较小的那个数) 

代码:

 1 #include <algorithm>
 2 #include <iostream>
 3 #include <sstream>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <cstdio>
 7 #include <string>
 8 #include <bitset>
 9 #include <vector>
10 #include <queue>
11 #include <stack>
12 #include <cmath>
13 #include <list>
14 //#include <map>
15 #include <set>
16 using namespace std;
17 /***************************************/
18 #define ll long long
19 #define int64 __int64
20 #define PI 3.1415927
21 /***************************************/
22 const int INF = 0x7f7f7f7f;
23 const double eps = 1e-8;
24 const double PIE=acos(-1.0);
25 const int d1x[]= {0,-1,0,1};
26 const int d1y[]= {-1,0,1,0};
27 const int d2x[]= {0,-1,0,1};
28 const int d2y[]= {1,0,-1,0};
29 const int fx[]= {-1,-1,-1,0,0,1,1,1};
30 const int fy[]= {-1,0,1,-1,1,-1,0,1};
31 const int dirx[]= {-1,1,-2,2,-2,2,-1,1};
32 const int diry[]= {-2,-2,-1,-1,1,1,2,2};
33 /*vector <int>map[N];map[a].push_back(b);int len=map[v].size();*/
34 /***************************************/
35 void openfile()
36 {
37     freopen("data.in","rb",stdin);
38     freopen("data.out","wb",stdout);
39 }
40 priority_queue<int> qi1;
41 priority_queue<int, vector<int>, greater<int> >qi2;
42 /**********************华丽丽的分割线,以上为模板部分*****************/
43 
44 int main()
45 {
46     //zuhe();
47     long long n,m;
48     while(scanf("%lld%lld",&n,&m)&&(n+m))
49     {
50         long long tt,t;
51         if (m<n)
52         {
53             t=n;
54             n=m;
55             m=t;
56         }
57         t=n;
58         tt=m+n;
59         double ce=1.0;
60         while(t)
61         {
62             ce=(double)(ce*tt)/(double)t;
63             t--;
64             tt--;
65         }
66         printf("%.0lf
",ce);
67     }
68     return 0;
69 }
View Code
原文地址:https://www.cnblogs.com/ZhaoPengkinghold/p/4002706.html