codeforces 690D2 D2. The Wall (medium)(组合数学)

题目链接：

Heidi the Cow is aghast: cracks in the northern Wall? Zombies gathering outside, forming groups, preparing their assault? This must not happen! Quickly, she fetches her HC2 (Handbook of Crazy Constructions) and looks for the right chapter:

How to build a wall:

Take a set of bricks.
Select one of the possible wall designs. Computing the number of possible designs is left as an exercise to the reader.
Place bricks on top of each other, according to the chosen design.

This seems easy enough. But Heidi is a Coding Cow, not a Constructing Cow. Her mind keeps coming back to point 2b. Despite the imminent danger of a zombie onslaught, she wonders just how many possible walls she could build with up to n bricks.

A wall is a set of wall segments as defined in the easy version. How many different walls can be constructed such that the wall consists of at least 1 and at most n bricks? Two walls are different if there exist a column c and a row r such that one wall has a brick in this spot, and the other does not.

Along with n, you will be given C, the width of the wall (as defined in the easy version). Return the number of different walls modulo106 + 3.

Input

The first line contains two space-separated integers n and C, 1 ≤ n ≤ 500000, 1 ≤ C ≤ 200000.

Output

Print the number of different walls that Heidi could build, modulo 10^6 + 3.

Examples

input

5 1

output

input

2 2

output

input

3 2

output

input

11 5

output

input

37 63

output

230574


题意;

给n个砖,给定了宽c,问你有多少种墙;
这也是[1,n]的按顺序拆分成c个数的种数;也是把[1,n]放在c个盒子里(允许有空盒)的方案数;

思路：

dp[n][m]=dp[n][m-1]+dp[n-1][m]+...+dp[0][m]+dp[n-1][m]+dp[n-2][m]+...+dp[1][m];
把n个相同的球放在m个盒子里(允许有空盒)的方案数为C(n+m-1,m-1);
dp[n][m]=C(n+m-1,m-1)=(n+m-1)/n*C(n+m-2,m-1)=(n+m-1)/n*dp[n-1][m];
取模的时候要快速幂求一下逆;

AC代码：

#include <bits/stdc++.h>
/*
#include <vector>
#include <iostream>
#include <queue>
#include <cmath>
#include <map>
#include <cstring>
#include <algorithm>
#include <cstdio>
*/
using namespace std;
#define For(i,j,n) for(int i=j;i<=n;i++)
#define mst(ss,b) memset(ss,b,sizeof(ss));
typedef  long long LL;
template<class T> void read(T&num) {
    char CH; bool F=false;
    for(CH=getchar();CH<'0'||CH>'9';F= CH=='-',CH=getchar());
    for(num=0;CH>='0'&&CH<='9';num=num*10+CH-'0',CH=getchar());
    F && (num=-num);
}
int stk[70], tp;
template<class T> inline void print(T p) {
    if(!p) { puts("0"); return; }
    while(p) stk[++ tp] = p%10, p/=10;
    while(tp) putchar(stk[tp--] + '0');
    putchar('
');
}

const LL mod=1e6+3;
const double PI=acos(-1.0);
const LL inf=1e18;
const int N=5e5+10;
const int maxn=1005;
const double eps=1e-10;


LL n,m,dp[N];

LL pow_mod(LL x,LL y)
{
    LL s=1,base=x;
    while(y)
    {
        if(y&1)s=s*base%mod;
        base=base*base%mod;
        y>>=1;
    }
    return s;
}

void Init()
{
    dp[0]=1;
    for(LL i=1;i<=n;i++)
    {
        dp[i]=(dp[i-1]*(i+m-1)%mod)*pow_mod(i,mod-2)%mod;
    }
}


int main()
{
        read(n);read(m);
        Init();
        LL sum=0;
        for(LL i=1;i<=n;i++)
        {
            sum=(sum+dp[i])%mod;
        }
        cout<<sum<<"
";

        return 0;
}