Hopscotch

问题 H: Hopscotch

题目描述

You’re playing hopscotch! You start at the origin and your goal is to hop to the lattice point (N, N). A hop consists of going from lattice point (x1, y1) to (x2, y2), where x1 < x2 and y1 < y2.
You dislike making small hops though. You’ve decided that for every hop you make between two lattice points, the x-coordinate must increase by at least X and the y-coordinate must increase by at least Y .
Compute the number of distinct paths you can take between (0, 0) and (N, N) that respect the above constraints. Two paths are distinct if there is some lattice point that you visit in one path which you don’t visit in the other.
Hint: The output involves arithmetic mod 109+ 7. Note that with p a prime like 109+ 7, and x an integer not equal to 0 mod p, then x(xp−2) mod p equals 1 mod p.
输入
The input consists of a line of three integers, N X Y . You may assume 1 ≤ X, Y ≤ N ≤ 106.
输出
The number of distinct paths you can take between the two lattice points can be very large. Hence output this number modulo 1 000 000 007 (109+ 7).

样例输入

7 2 3

样例输出

9

题意

给出一个n*n的网格，从(0,0)点走到(n,n)有多少种走法，其中每次走x方向最少走x个格子y方向最少走y个格子

题解

将x和y两个方向分开来看，记录两个数组dp1,dp2，其中dp1[i]代表x方向上从0走到用i步有多少种走法，dp2[i]代表y方向上
其中dp1,dp2的值可以用组合数学求得，每次给i步分配（x-1）个格子，然后将剩下的格子用隔板法分配给各个步就可以了；
最后再将两个dp求积，然后求一下和就是答案了。

#include<cstdio>
#include<vector>
#include<algorithm>
#include <iostream>
typedef long long ll;
using namespace std;
ll mod = 1000000007;
const int maxn = 1000010;

ll qpow(ll a,ll x){
    ll ret=1;
    while (x){
        if (x&1)
            ret = ret*a%mod;
        a=a*a%mod;
        x>>=1;
    }
    return ret;
}
ll fac[maxn],inv[maxn];

ll init(){
    fac[0]=1;
    for (int i=1;i<maxn;i++)
        fac[i]=fac[i-1]*i%mod;
    inv[maxn-1]=qpow(fac[maxn-1],mod-2);
    for (int i=maxn-2;i>=0;i--)
        inv[i]=inv[i+1]*(i+1)%mod;
    return 0;
} 
ll c(ll n,ll m){
    if (n<m) return 0;
    return fac[n]*inv[m]%mod*inv[n-m]%mod;
}
ll dp1[maxn],dp2[maxn];
int main() 
{ 
    ll n,x,y;
    scanf("%lld%lld%lld",&n,&x,&y); 
    for (int i=1;i*x<=n;i++)
    {
        dp1[i]=c(n-(x-1)*i-1,i-1);
    }
    for (int i=1;i*y<=n;i++)
    {
        dp2[i]=c(n-(y-1)*i-1,i-1);
    }
    ll ans = 0;
    for (int i=1;i*x<=n&&i*y<=n;i++)
        ans =( ans + dp1[i]*dp2[i]%mod)%mod ;
    printf("%lld\n",ans);
    return 0;
}