HEX

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 13 2100000 100000

Example Output

131

Hint

存个代码，提示就是选路问题，有三个走法，下，下左，下右。

用组合数学做

#include <bits/stdc++.h>using namespace std;const long long mod = 1e9+7;const int maxn = 1e5+10;typedef long long LL;LL fac[maxn],Ni[maxn];LL modxp(LL a,LL b,LL p){    LL res = 1;    while(b){        if(b&1)            res = (res*a)%p;        b = b>>1;        a = (a*a)%p;    }    return res;}LL niYuan(LL a, LL b){//费马小定理求逆元    return modxp(a, b - 2, b);}void Init(){    fac[0] = 1;    Ni[0] = 1;    for(int i = 1; i < maxn; i++){        fac[i] = (fac[i-1]*i)%mod;        Ni[i] = niYuan(fac[i],mod);    }}LL C(LL a,LL b){    return ((fac[a]*Ni[b])%mod*Ni[a-b]% mod)%mod;}int main(){    LL x,y,ans,a,b,c,pian;    Init();    while(~scanf("%lld %lld",&y,&x)){        ans = 0;        if(y%2 == 0){            if(x>=y/2+1)                pian = (x-(y/2+1))*2+1;            else                pian = (y/2-x)*2+1;        }        else{            pian = abs((y+1)/2-x)*2;        }        a = pian;        b = 0;        while(a+b<=y-1){            c = (y-1-a-b)/2;            ans = ans+(C(a+b+c,a)*C(b+c,b))%mod;            ans = ans%mod;            a++;            b++;        }        printf("%lld\n",ans);    }    return 0;}