山东 acm D HEX 数学 求组合数

HEX
Time Limit: 4000MS Memory Limit: 131072KB
Submit Statistic
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.
Example Input

1 1
3 2
100000 100000
Example Output

1
3
1
Hint

Author

利用 （1,1）+x*（1,0）+y*（2,1）+z*（1,1）=（a,b）.
求得 x+2*y+z=a-1;y+z=b-1; 得 x+y=a-b; y+z=b-1 y<=min(a-b,b-1)

枚举y，解出，x，z，然后解出 c（x+y+z，x）*C(y+z,y)*C(z,z) ，表明走x个左方向 y个下方向和z个右方向有多少种方案, 用 res加上。

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;ll fact[101010],inv[101010];const int mod=1e9+7;const int maxn=1e5+7;ll quickM(ll a, ll b){    ll base = a%mod, ans = 1;    while(b){        if(b&1) ans = ans*base%mod;        base = base*base%mod;        b >>= 1;     }    return ans;}void init(){    fact[0] = 1;    for(int i = 1; i <= maxn; ++i)    fact[i] = fact[i-1]*i%mod;    inv[maxn]=quickM(fact[maxn],mod-2);    for(int i=maxn-1;i>=0;i--)    {        inv[i]=inv[i+1]*(i+1);        inv[i]%=mod;    }}ll C(int n, int m){    return ((fact[n]*inv[m])%mod*(inv[n-m]))%mod;}int main(){    int a,b;    init();    while(cin>>a>>b)    {        long long res=0;        int n=min(a-b,b-1);        for(ll y=0;y<=min(a-b,b-1);y++)        {            ll z=b-1-y;            ll x=a-b-y;            res+=(C(z+y+x,x)*C(z+y,y))%mod;            res%=mod;        }        printf("%lld\n",res );    }}