2016暑假多校联合---A Simple Chess
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.
(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).
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 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
思路：先找马可以走到的点，可以发现点均分布在斜率为-1的直线上，个数为1、2、3、4…… 所以可以把这些点对应到第一象限，这些点分别为（1,1） （1,2）、（2,1）
（1,3）、（2,2）、（3,1） （1,4）（2,3）、（3,2）、（4,1） 刚好充满第一象限，这时可以用组合数方便算出从某点到另一点的路径数，由于坐标很大，所以必须使用卢卡斯定理求组合数，注意去重，因为有障碍点。
#include <iostream> #include <algorithm> #include <stdio.h> #include <string.h> #include <cmath> using namespace std; const long long mod=110119; typedef long long LL; struct Node { long long x; long long y; long long s; }node[105]; int cmp(const Node s1,const Node s2) { if(s1.x==s2.x) return s1.y<s2.y; return s1.x<s2.x; } LL PowMod(LL a,LL b,LL MOD) { LL ret=1; while(b) { if(b&1) ret=(ret*a)%MOD; a=(a*a)%MOD; b>>=1; } return ret; } LL fac[1000005]; LL Get_Fact(LL p) { fac[0]=1; for(int i=1; i<=p; i++) fac[i]=(fac[i-1]*i)%p; } LL calc(LL n,LL m,LL p) { LL ret=1; while(n&&m) { LL a=n%p,b=m%p; if(a<b) return 0; ret=(ret*fac[a]*PowMod(fac[b]*fac[a-b]%p,p-2,p))%p; n/=p; m/=p; } return ret; } int main() { long long n,m; int r; int Case=1; Get_Fact(mod); while(scanf("%lld%lld%d",&n,&m,&r)!=EOF) { int tot=0; long long sum=0; int flag=0; if((n+m+1)%3==0) { long long s=(n+m+1)/3; if(n>=s&&m>=s) { long long t=n; n=2*s-m; m=2*s-t; } else { flag=1; } } else { flag=1; } for(int i=0;i<r;i++) { long long x,y; scanf("%lld%lld",&x,&y); if((x+y+1)%3==0) { long long s=(x+y+1)/3; if(x>=s&&y>=s) { node[tot].x=2*s-y; node[tot].y=2*s-x; if(node[tot].x<=n&&node[tot].y<=m) tot++; } } } if(flag==1) { printf("Case #%d: %lld\n",Case++,sum); continue; } if(tot>0) sort(node,node+tot,cmp); sum=calc(n+m-2,n-1,mod)%mod; // cout<<"n: "<<n<<" m: "<<m<<endl; //cout<<tot<<endl; //cout<<sum<<endl; //for(int i=0;i<tot;i++) //cout<<"tot: "<<node[i].x<<" "<<node[i].y<<endl; for(int i=0;i<tot;i++) { node[i].s=calc(node[i].x+node[i].y-2,node[i].x-1,mod)%mod; } for(int i=0;i<tot;i++) { long long tt=calc(n-node[i].x+m-node[i].y,m-node[i].y,mod); for(int j=i+1;j<tot;j++) { if(node[j].y>=node[i].y) { long long d1=node[j].y-node[i].y; long long d2=node[j].x-node[i].x; node[j].s=(node[j].s-(node[i].s*calc(d1+d2,d1,mod))%mod)%mod; } } sum=(sum-(node[i].s*tt)%mod)%mod; sum = (sum%mod+mod)%mod; } printf("Case #%d: %lld\n",Case++,sum); } }
