Hopscotch

时间限制: 5 Sec 内存限制: 128 MB

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

题解

题目大意: 给定三个数n,x,y。表示有一个n*n的方阵，如果x1+x<x2且y1+y<y2，那么可以从(x1,y1)跳到(x2,y2)，问从(0,0)跳到(n,n)有多少种不同的走法。
先把行列分开考虑。
a1[i]表示一列中，从0走到n使用了i步有多少组走法，a2[i]表示一行中，从0走到n使用了i步有多少组走法，则使用i步从(0,0)走到(n,n)的方法数为a1[i]*a2[i],所以答案ans = a1[1]*a2[1]+a1[2]*a2[2]+...a1[i]*a2[i]。
那么如何求出两个a数组呢？我们可以这样考虑，一共有n个点，走i次，这样相当于把n个连续的小球分到i个桶里。因为每个桶至少分x个球，所以我们可以先给每个桶分x-1个球，剩下的n-(x-1)*i个球用隔板法分给i个人。也就是a1[i]=C(n-(x-1)*i-1,i-1)种。

代码

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

ll fac[maxn],inv[maxn];
ll a1[maxn],a2[maxn];

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 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;
}

int main() {
    init();
    ll n, x, y;
    scanf("%lld%lld%lld", &n, &x, &y);
    for (int i = 1; i * x <= n; i++) {
        a1[i] = c(n - (x - 1) * i - 1, i - 1);
    }
    for (int i = 1; i * y <= n; i++) {
        a2[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 + a1[i] * a2[i] % mod) % mod;
    printf("%lld\n", ans);
    return 0;
}
/**************************************************************
    Problem: 5095
    User: DP56
    Language: C++
    Result: 正确
    Time:916 ms
    Memory:32948 kb
****************************************************************/