HEX
Problem Description
On a plain of hexagonal grid, we define a step as one move from the current grid to the lower/lower-left/lower-right grid. For example, we can move from (1,1) to (2,1), (2,2) or (3,2).
In the following graph we give a demonstrate of how this coordinate system works.
Your task is to calculate how many possible ways can you get to grid(A,B) from gird(1,1), where A and B represent the grid is on the B-th position of the A-th line.
Input
For each test case, two integers A (1<=A<=100000) and B (1<=B<=A) are given in a line, process till the end of file, the number of test cases is around 1200.
Output
For each case output one integer in a line, the number of ways to get to the destination MOD 1000000007.
Sample Input
1 1 3 2 100000 100000
Sample Output
1 3 1
Hint
Source
"Inspur Cup" Shandong Province 8th ACM College Student Programming Competition (Thanks to Qingdao University of Science and Technology)
Idea: If you can only go to ↙ or ↘, from (1,1)->(x,y), you have to take tot=(x-1) steps in total, and r=(y-1) steps to ↘ , to take l=(xy) steps towards ↙, then the total move is C(tot,r) (or C(tot,l)). If you can still go to ↓, because one step ↓ is equivalent to one step ↙ plus one step ↘, so if you take i steps to ↓, In total, just take tot'=(tot-i) steps, r'=(ri) steps towards ↘, and l'=(li) steps towards ↙, (where 0<=i<=min(l,r) ), Then the total number of moves is C(tot',i)+C(tot'-i,r) (more than one expression).
code:
#include <bits/stdc++.h> using namespace std; const int mod = 1e9+7; typedef long long ll; ll fac[110000]; void init(){ fac [0] = fac [1] = 1; for(int i = 2; i <= 100000; i++){ fac [i] = (fac [i-1] * i)% mod; } } ll q_pow(ll a,ll b){ ll ans = 1; while(b){ if(b & 1) years = years * a % mod; b >>= 1; a = a * a % mod; } return ans % mod; } ll C(ll n,ll m){ if(m > n) return 0; return fac [n] * q_pow (fac [m] * fac [nm], mod - 2)% mod; } ll Lucas(ll n,ll m){ if(m == 0) return 1; else return (C(n%mod,m%mod) * Lucas(n/mod,m/mod)) % mod; } int main(){ init(); ll x,y; while(~scanf("%lld%lld",&x,&y)){ ll all = x - 1; ll r1 = y - 1; ll l1 = x - y; ll down = min(l1,r1); ll ans = 0; for(ll i = 0; i <= down; i++){ ll tot = all - i; ll l = l1 - i; ll r = r1 - i; ans = (ans + Lucas (tot, i)% mod * Lucas (tot-i, l)% mod)% mod; } printf("%lld\n",ans); } return 0; }