您的位置：首页 > 其它

HDU 5794 A Simple Chess dp+Lucas

2016-08-10 01:10 513 查看

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=5794

A Simple Chess

Time Limit: 2000/1000 MS (Java/Others)Memory Limit: 65536/65536 K (Java/Others)

问题描述

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.

输入

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).

输出

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.

样例

sample input

1 1 0

3 3 0

4 4 1

2 1

4 4 1

3 2

7 10 2

1 2

7 1

sample output

Case #1: 1

Case #2: 0

Case #3: 2

Case #4: 1

Case #5: 5

题意

n*m的大棋盘，有r个障碍物，你在1,1,并且每次只能往左下走日子步，问到达(n,m)的方案数

题解

Lucas+dp.

预备：设从i点到j点过程中走横日走了x步，走竖日走了y步，则有方程2x+y==n&&x+2y==m --x=(2n-m)/3,y=(2m-n)/3。

所以方案数为C[x+y][x]。这个需要用卢卡斯处理出来。

先把所有障碍点从左上到右下排序。

dp[i]表示从1,1到(pt[i].x,pt[i].y)障碍的不经过任意其他位于它左上的障碍的情况数，则dp[i]=dp[i]-sigma(dp[j]*j到i点的所有方案数）。

代码

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define X first
#define Y second
#define mkp make_pair
#define lson (o<<1)
#define rson ((o<<1)|1)
#define mid (l+(r-l)/2)
#define pb(v) push_back(v)
#define sz() size()
#define all(o) (o).begin(),(o).end()
#define clr(a,v) memset(a,v,sizeof(a))
#define bug(a) cout<<#a<<" = "<<a<<endl
#define rep(i,a,b) for(int i=a;i<(b);i++)

typedef long long  LL;
typedef vector<int> VI;
typedef pair<int,int> PII;
typedef vector<pair<int,int> > VPII;

const int INF=0x3f3f3f3f;
const LL INFL=0x3f3f3f3f3f3f3f3fLL;
const double eps=1e-8;
const double PI=acos(-1.0);

//start----------------------------------------------------------------------

const int maxn=111;
const int mod=110119;
const int maxm=mod+10;

LL n,m; int r;
pair<LL,LL> pt[maxn];
LL facinv[maxm],inv[maxm],fac[maxm];
LL dp[maxn];

LL get_C(LL n,LL m){
if(n<0||m<0||n<m) return 0;
return fac
*facinv[m]%mod*facinv[n-m]%mod;
}

LL Lucas(LL n,LL m,int mod){
if(m==0) return 1LL;
return get_C(n%mod,m%mod)*Lucas(n/mod,m/mod,mod)%mod;
}

LL calc(int i,int j){
LL n=pt[j].X-pt[i].X;
LL m=pt[j].Y-pt[i].Y;
if((n+m)%3) return 0;
LL sum=(n+m)/3;
if(n<0||m<0) return 0;
return Lucas(sum,n-sum,mod);
}

void pre(){
fac[0]=fac[1]=1;
facinv[0]=facinv[1]=1;//facinv[0]=1!!!!
inv[1]=1;
rep(i,2,maxm){
fac[i]=fac[i-1]*i%mod;
inv[i]=(mod-mod/i)*inv[mod%i]%mod;
facinv[i]=facinv[i-1]*inv[i]%mod;
}
}

int main(){
pre();
int kase=0;
while(scanf("%lld%lld%d",&n,&m,&r)==3){
clr(dp,0);
int flag=0;
rep(i,1,r+1){
scanf("%lld%lld",&pt[i].X,&pt[i].Y);
if(pt[i].X==n&&pt[i].Y==m){
flag=1;
}
}
if(flag){
printf("Case #%d: 0\n",++kase);
continue;
}
pt[0].X=1,pt[0].Y=1;
pt[r+1].X=n,pt[r+1].Y=m;

sort(pt+1,pt+r+1);
dp[0]=1;
rep(i,1,r+2){
dp[i]=calc(0,i);
rep(j,1,i){
dp[i]-=dp[j]*calc(j,i)%mod;
dp[i]=(dp[i]%mod+mod)%mod;
}
}
printf("Case #%d: %lld\n",++kase,dp[r+1]);
}
return 0;
}

//end-----------------------------------------------------------------------

内容来自用户分享和网络整理，不保证内容的准确性，如有侵权内容，可联系管理员处理

标签：

相关文章推荐

新的分享

章节导航