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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 long19 #define int64 __int6420 #define PI 3.141592721 /***************************************/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\n",ce);67     }68     return 0;69 }
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poj1942(Paths on a Grid)