bzoj4403题解

【参考代码】

#pragma GCC optimize(2)
#include <cstdlib>
#define function(type) __attribute__((optimize("-O2"))) inline type
#define procedure __attribute__((optimize("-O2"))) inline void
using namespace std;

//quick_io {
#include <cctype>
#include <cstdio>
function(long long) getint()
{
	char c=getchar(); for(;!isdigit(c)&&c!='-';c=getchar());
	short s=1; for(;c=='-';c=getchar()) s*=-1; long long r=0;
	for(;isdigit(c);c=getchar()) r=(r<<3)+(r<<1)+c-'0';
	return s*r;
}
//} quick_io

static const int AwD=1000003;

function(int) format(const int&x) {return x+(x<0)*AwD;}

function(int) mrev(const int&x)
{
	int ret=1,bas=x;
	for(register int i=AwD-2;i;i>>=1,bas=1ll*bas*bas%AwD)
	{
		if(i&1) ret=1ll*ret*bas%AwD;
	}
	return ret;
}

int fact[AwD]={1};

function(int) Lucas(const int&m,const int&n)
{
	div_t M=div(m,AwD),N=div(n,AwD);
	if(M.quot<N.quot||M.rem<N.rem) return 0;
	return 1ll*fact[M.quot]*fact[M.rem]*mrev(
		1ll*fact[N.quot]*fact[N.rem]%AwD*fact[M.quot-N.quot]*fact[M.rem-N.rem]%AwD
	)%AwD;
}

int main()
{
	for(register int i=1;i<AwD;++i) fact[i]=1ll*fact[i-1]*i%AwD;
	for(register int T=getint();T--;)
	{
		int n=getint(),l=getint(),r=getint();
		printf("%d\n",format(Lucas(n-l+r+1,n)-1));
	}
	return 0;
}