HDU5794-A Simple Chess（Lucas定理）

A Simple Chess

                                                                    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
                                                                                            Total Submission(s): 2905    Accepted Submission(s): 764

Problem Description

There is a n×m board, a chess want to go to the position 
(n,m) from the position (1,1).
The chess is able to go to position (x2,y2) from the position (x1,y1), only and if only x1,y1,x2,y2 is satisfied that (x2−x1)2+(y2−y1)2=5, x2>x1, y2>y1.
Unfortunately, there are some obstacles on the board. And the chess never can stay on the grid where has a obstacle.
I want you to tell me, There are how may ways the chess can achieve its goal.

 

Input

The input consists of multiple test cases.
For each test case:
The first line is three integers, n,m,r,(1≤n,m≤1018,0≤r≤100), denoting the height of the board, the weight of the board, and the number of the obstacles on the board.
Then follow r lines, each lines have two integers, x,y(1≤x≤n,1≤y≤m), denoting the position of the obstacles. please note there aren't never a obstacles at position (1,1).

 

Output

For each test case,output a single line "Case #x: y", where x is the case number, starting from 1. And y is the answer after module 110119.

 

Sample Input

1 1 03 3 04 4 12 14 4 13 27 10 21 27 1

 

Sample Output

Case #1: 1Case #2: 0Case #3: 2Case #4: 1Case #5: 5

 

Author

UESTC

 

Source

2016 Multi-University Training Contest 6

 

Recommend

wange2014

 

题意：给你一个n*m的网格，问从(1, 1)走到(n, m)的方案数是多少，其中有r个障碍，障碍不能走，每次只能走”日"型

解题思路：可达点都是满足(x + y) % 3 = 2的; 路径可以看成一个斜着放置的杨辉三角。我们只需要把坐标转换一下即可，这是没有障碍时的方案数，有一个障碍，那么可以用起点到终点的方法数 - 起点到障碍点的方法数*障碍点到终点的方法数

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <algorithm>#include <map>#include <cmath>#include <set>#include <stack>#include <queue>#include <vector>#include <bitset>#include <functional>using namespace std;#define LL long longconst int INF = 0x3f3f3f3f;#define mod 110119struct node{LL x, y;bool operator < (const node &a)const{if (x != a.x) return x < a.x;return y < a.y;}}a[105];LL f[120000], ans[105];void Get_Fact(){f[0] = 1;for (int i = 1; i <= mod; i++) f[i] = (f[i - 1] * i) % mod;}LL Pow(LL a, LL b){LL ans = 1;while (b){if (b & 1) ans = ans*a%mod;b >>= 1;a = a*a%mod;}return ans;}LL C(LL n, LL m){if (m > n) return 0;if (m == 0) return 1;LL ans = f[n] * Pow(f[m], mod - 2) % mod * Pow(f[n - m], mod - 2) % mod;return ans;}LL Lucas(LL n, LL m){if (n < 0 || m < 0) return 0;if (m > n) return 0;if (m == 0) return 1;return C(n%mod, m%mod) * Lucas(n / mod, m / mod) % mod;}LL solve(LL x1, LL y1, LL x2, LL y2){if ((x1 + y1) % 3 != 2) return 0;if ((x2 + y2) % 3 != 2) return 0;LL ax = (x1 + y1 - 2) / 3;LL ay = y1 - 1 - ax;LL bx = (x2 + y2 - 2) / 3;LL by = y2 - 1 - bx;return Lucas(bx - ax, by - ay);}int main(){Get_Fact();LL n, m;int cas = 1, r;while (~scanf("%lld %lld %d", &n, &m, &r)){for (int i = 1; i <= r; i++) scanf("%lld %lld", &a[i].x, &a[i].y);sort(a + 1, a + r + 1);LL sum = solve(1, 1, n, m);for (int i = 1; i <= r; i++){ans[i] = solve(1, 1, a[i].x, a[i].y);for (int j = 1; j < i; j++)ans[i] = (ans[i] - ans[j] * solve(a[j].x, a[j].y, a[i].x, a[i].y) % mod + mod) % mod;}for (int i = 1; i <= r; i++) sum = (sum - ans[i] * solve(a[i].x, a[i].y, n, m) % mod + mod) % mod;printf("Case #%d: %lld\n", cas++, sum);}return 0;}