C++14(g++5.4) 解法, 执行用时: 222ms, 内存消耗: 18660K, 提交时间: 2020-06-12 14:05:24

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod = 911451407;
const int maxn = 2e6 + 10;
ll mul[maxn];
ll q_pow(ll a,ll b)
{
	ll res = 1;
	while(b)
	{
		if(b&1)res = res*a%mod;
		a = a*a%mod;
		b>>=1;
	}
	return res;
}
ll C(ll n,ll m)
{
	if(m==0)return 1;
	return mul[n]*q_pow(mul[n - m]*mul[m]%mod,mod - 2)%mod;
	 
}
int main()
{
	int T;
	ll l,x,num,cnt,ans;
	cin >> T;
	mul[0] = 1;
	for(ll i = 1;i < maxn;i++)
	mul[i] = mul[i - 1]*i%mod;
	while(T--)
	{
		scanf("%lld %lld",&l,&x);
		cout << C(x + l - 2,l - 1) << endl;
	}
	return 0;
}

C++(g++ 7.5.0) 解法, 执行用时: 374ms, 内存消耗: 18408K, 提交时间: 2023-04-01 22:08:26

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
ll f[(int)1e6*2+53];
const ll mod=911451407;
ll ksm(ll a,ll b){
    ll ans=1;
    while(b){
       if(b&1){ans=(ans*a)%mod;}
        b>>=1;a=(a*a)%mod;
    }
    return ans;
}
int main(){
    int t,l,x;
    f[0]=1;
    for(int i=1;i<=1e6*2+3;i++){
        f[i]=f[i-1]*i;
        f[i]%=mod;
    }
    cin>>t;
    while(t--){
        cin>>l>>x;
        cout<<f[x+l-2]*ksm(f[l-1],mod-2)%mod*ksm(f[x-1],mod-2)%mod<<"\n";
    }
    return 0;
}

C++11(clang++ 3.9) 解法, 执行用时: 176ms, 内存消耗: 32748K, 提交时间: 2020-06-05 20:21:19

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e6+5;
const int mod=911451407;
ll f[N],invf[N];
ll C(ll n,ll m)
{
    return f[n]*invf[n-m]%mod*invf[m]%mod;
}
int main()
{
    f[0]=f[1]=invf[0]=invf[1]=1;
    for(int i=2;i<N;i++)f[i]=f[i-1]*i%mod,invf[i]=(mod-mod/i)*invf[mod%i]%mod;
    for(int i=2;i<N;i++)invf[i]=invf[i]*invf[i-1]%mod;
    int t;scanf("%d",&t);
    while(t--)
    {
        int l,x;scanf("%d%d",&l,&x);
        printf("%lld\n",C(l-1+x-1,x-1));
    }
}