省SD2017 D HEX【乘法逆元+排列+转化】

HEX

Time Limit: 4 Sec Memory Limit: 128 MB
Submit: 6 Solved: 4
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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

山东省第八届ACM程序设计大赛2017.5.7

#include<iostream>
#include<cstdio>
using namespace std;
#define ll long long int 
#define mod 1000000007
ll i, x, y, a, b, c, ans;
ll p[112345];
ll n[112345];
ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (!b)
    {
        x = 1, y = 0;
        return a;
    }
    ll r = exgcd(b, a%b, x, y);
    ll temp = x;
    x = y;
    y = temp - a / b*y;
    return r;
}
ll cal(ll a, ll m)
{
    ll ans, x, y;
    ll gcd = exgcd(a, m, x, y);
    if (1 % gcd)
        return -1;
    x *= 1 / gcd;
    m = abs(m);
    ans = x%m;
    if (ans < 0)
        ans += m;
    return ans;
}
int main()
{
    p[0] = 1;
    n[0] = cal(p[0], mod);
    for (i = 1;i < 112345; i++)
    {
        p[i] = (p[i - 1] * i) % mod;
        n[i] = cal(p[i], mod);
    }
    while (cin >> x >> y)
    {
        ans = c = 0;
        a = x - y;
        b = x - a - 1;
        while (~a&&~b)
        {
            ans = (ans + p[a + b + c] * n[a] % mod*n[b] % mod*n[c] % mod) % mod;
            a--, b--, c++;
        }
        cout << ans << endl;
    }
}
//ll x, y;

//ll C(ll n,ll m)
//{
//  ll ans = 1;
//  for (int i = m+1; i <= n; i++)
//  {
//      ans *= i;
//      ans %= MOD;
//  }
//  for (int i = 1; i <= n - m; i++)
//  {
//      ans /=i;
//      ans %= MOD;
//  }
//  return (ans + MOD) % MOD;
//}
//int main()
//{
//  while (cin >> x >> y)
//  {
//      x = x - y + 1;
//      x--;
//      y--;
//      ll l, s, r;//向左、向下、向右
//      ll answer = 1;
//      if (x == 0 || y == 0)
//      {
//          cout << "1" << endl;
//          continue;
//      }
//      for (int i = 0; i <= x; i++)
//      {
//          l = i;
//          s = x - i;
//          r = y - s;
//          answer += (C(x, i)*C(y , r)) % MOD;
//          answer %= MOD;
//      }
//      answer += MOD;
//      answer %= MOD;
//      cout << answer << endl;
//  }
//  return 0;
//}