[BZOJ4403]序列统计

题意：给定三个正整数N、L和R，统计长度在1到N之间，元素大小都在L到R之间的单调不降序列的数量。输出答案对10^6+3取模的结果。
我的数学好差啊。。。

推式子见：http://www.cnblogs.com/Var123/p/5546290.html

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<string>
#include<ctime>
#include<queue>
#include<map>
#include<set>
#include<vector>
typedef long long LL;
using namespace std;
const LL p=1000003;
LL T,n,R,L,fac[p+1];
LL read() {LL d=0,f=1; char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1; c=getchar();} while (c>='0'&&c<='9') d=(d<<3)+(d<<1)+c-48,c=getchar(); return d*f;}
void judge(){freopen(".in","r",stdin); freopen(".out","w",stdout);}
LL quickmi(LL a,LL b)
{
    LL res=1;
    while (b)
    {
        if (b&1) res=res*a%p;
        a=a*a%p; b>>=1;
    }
    return res;
}
LL C(LL n,LL m)
{
    if (n<m) return 0;
    if (n==m) return 1;
    return fac[n]*quickmi(fac[m]*fac[n-m]%p,p-2)%p;
}
LL lucas(LL n,LL m)
{
    LL res=1;
    while (n&&m&&res)
    {
        res=res*C(n%p,m%p)%p;
        n/=p; m/=p;
    }
    return res;
}
int main()
{
    //judge();
    fac[0]=1;
    for (int i=1;i<=p;i++) fac[i]=fac[i-1]*i%p;
    T=read();
    while (T--)
    {
        n=read(); L=read(); R=read();
        LL ans=lucas(n+R-L+1,R-L+1);
        printf("%lld\n",(ans-1+p)%p);
    }
    return 0;
}