nCr%m when m is not prime and n and r is sufficiently large.

In many problems we need to calculate nCr%m where n, r and m are three positive integers. If the mod value m is a prime number then we can calculate nCr%m in different ways like using loops, using pascal’s triangle, using modular multiplicative inverse, using dp technique etc. This ways are described with source codes in this post.

Now our problem arrive when the mod value m is not prime. In this case we can’t use the above techniques. In this case we need to use the chinese remainder theorem (CRT) and Andrew Granville’s theory for calculating nCr. Here I provide you some ways to learn this techniques. I think this ways will be helpful to you.

1. First learn about chinese remainder theorem (CRT). You can learn it form these sources-
a. geeksforgeeks 1
b. geeksforgeeks 2
c. youtube 1
d. youtube 2

2. Second you can have a look on Andres Granville’s theory. The theory is explained here.

3. Now you can have a look on this problem. The detailed algorithm for our job is explained in this problem’s editorial section.

4. Now you can try to implement the algorithm. If you find any difficulties after several tries then you can see my implementation. Which is given below.


#include <bits/stdc++.h>

#define pii              pair <int,int>
#define pll              pair <long long,long long>
#define sc               scanf
#define pf               printf
#define Pi               2*acos(0.0)
#define ms(a,b)          memset(a, b, sizeof(a))
#define pb(a)            push_back(a)
#define MP               make_pair
#define db               double
#define ll               long long
#define EPS              10E-10
#define ff               first
#define ss               second
#define sqr(x)           (x)*(x)
#define D(x)             cout<<#x " = "<<(x)<<endl
#define VI               vector <int>
#define DBG              pf("Hi\n")
#define MOD              1000000007
#define CIN              ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define SZ(a)            (int)a.size()
#define sf(a)            scanf("%d",&a)
#define sfl(a)           scanf("%lld",&a)
#define sff(a,b)         scanf("%d %d",&a,&b)
#define sffl(a,b)        scanf("%lld %lld",&a,&b)
#define sfff(a,b,c)      scanf("%d %d %d",&a,&b,&c)
#define sfffl(a,b,c)     scanf("%lld %lld %lld",&a,&b,&c)
#define stlloop(v)       for(__typeof(v.begin()) it=v.begin();it!=v.end();it++)
#define loop(i,n)        for(int i=0;i<n;i++)
#define loop1(i,n)       for(int i=1;i<=n;i++)
#define REP(i,a,b)       for(int i=a;i<b;i++)
#define RREP(i,a,b)      for(int i=a;i>=b;i--)
#define TEST_CASE(t)     for(int z=1;z<=t;z++)
#define PRINT_CASE       printf("Case %d: ",z)
#define LINE_PRINT_CASE  printf("Case %d:\n",z)
#define CASE_PRINT       cout<<"Case "<<z<<": "
#define all(a)           a.begin(),a.end()
#define intlim           2147483648
#define infinity         (1<<28)
#define ull              unsigned long long
#define gcd(a, b)        __gcd(a, b)
#define lcm(a, b)        ((a)*((b)/gcd(a,b)))

using namespace std;


//----------------------Graph Moves----------------
//const int fx[]={+1,-1,+0,+0};
//const int fy[]={+0,+0,+1,-1};
//const int fx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const int fy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const int fx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const int fy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
//------------------------------------------------

//-----------------------Bitmask------------------
//int Set(int N,int pos){return N=N | (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
//------------------------------------------------

#define mx 1000006

bitset<mx/2>vis;
vector<int>prime;

vector<pii>factor;

void sieve()
{
    int x=mx/2,y=sqrt(mx)/2;
    for(int i=1; i<=y; i++)
    {
        if(vis[i]==0)
        {
            for(int j=i*(i+1)*2; j<x; j+=(2*i)+1)
                vis[j]=1;
        }
    }

    prime.pb(2);

    for(int i=3; i<mx; i+=2)
        if(vis[i/2]==0)
            prime.pb(i);

}

ll factorial[mx];
ll arr[mx];


vector<ll>ans;

void precal(ll p, ll q, ll mod)
{
    arr[0]=1;
    arr[1]=1;
//    ll mod=bigmod(p,q,MOD);
    ll x=1;
    for(ll i=2; i<=mod; i++)
    {
        if(i%p)
            x=i;
        else
            x=1;
        arr[i]=(arr[i-1]*x)%mod;
    }
}

ll bigmod(ll n, ll p, ll mod)
{
    ll ret=1;
    while(p)
    {
        if(p%2)
            ret=(ret*n)%mod;
        n=(n*n)%mod;
        p/=2;
    }
    return ret;
}

ll E(ll n, ll p)
{
    ll ret=0;
    while(n)
    {
        ret+=n/p;
        n=n/p;
    }
    return ret;
}

ll f(ll n, ll mod)
{
    ll ret=bigmod(arr[mod-1],n/mod,mod)*arr[n%mod];
    return ret;
}

ll F(ll n, ll mod, ll p)
{
    ll ret=1;
    ll i=1;
    while(i<=n)
    {
        ret=(ret*f(n/i,mod))%mod;
        i=i*p;
    }
    return ret;
}

int inv(int a, int m) // Calculating Modular Multiplicative Inverse
{
    int m0 = m, t, q;
    int x0 = 0, x1 = 1;

    if (m == 1)
        return 0;

//     Apply extended Euclid Algorithm
    while (a > 1)
    {
//         q is quotient
        q = a / m;

        t = m;

//         m is remainder now, process same as
//         euclid's algo
        m = a % m, a = t;

        t = x0;

        x0 = x1 - q * x0;

        x1 = t;
    }

//     Make x1 positive
    if (x1 < 0)
        x1 += m0;

    return x1;
}




ll nCr(ll n, ll r, ll p, ll mod)
{
    ll e=E(n,p)-E(r,p)-E(n-r,p);
    ll mod1=F(n,mod,p);
    ll mod2=(F(r,mod,p)*F(n-r,mod,p))%mod;
    mod2=inv(mod2,mod);
    ll ret=bigmod(p,e,mod);
    ret*=mod1;
    ret%=mod;
    ret*=mod2;
    ret%=mod;
    return ret;
}

ll findMinX(int k) // Chinese Remainder
{
    ll prod = 1;
    vector<int>num;
    for(int i=0; i<k; i++)
    {
        num.pb(bigmod(factor[i].ff,factor[i].ss,MOD));
    }
    for (int i = 0; i < k; i++)
        prod *= num[i];

    ll result = 0;

    for (int i = 0; i < k; i++)
    {
        ll pp = prod / num[i];
        result += ans[i] * inv(pp, num[i]) * pp;
    }

    return result % prod;
}

ll nCr_mod_m(ll n, ll r, ll m)
{
    factor.clear();
    ans.clear();
    int root=sqrt(m);
    ll mm=m;
    for(int i=0; i<SZ(prime) && prime[i]<=root; i++)
    {
        if(mm%prime[i]==0)
        {
            int cnt=0;
            while(mm%prime[i]==0)
            {
                mm/=prime[i];
                cnt++;
            }
            factor.pb(pii(prime[i],cnt));
            root=sqrt(mm);
        }
    }

    if(mm>1)
        factor.pb(pii(mm,1));



    for(int i=0; i<SZ(factor); i++)
    {
        ll p=factor[i].ff;

        ll num=bigmod(p,factor[i].ss,MOD);
        precal(p,factor[i].ss,num);
        ans.pb(nCr(n,r,p,num));
    }

    ll anss=findMinX(SZ(factor));
    return anss;
}

int main()
{

//    freopen("in.txt","r",stdin);
//	  freopen("out.txt","w",stdout);

    sieve();

    int t;
    sf(t);
    TEST_CASE(t)
    {
        ll n,r,m;
        cin>>n>>r>>m;

        ll ans=nCr_mod_m(n,r,m);

        pf("%lld C %lld mod %lld = %lld\n",n,r,m,ans);
    }

    return 0;
}


Practice problems:
1. nCr
2. Codechef’s Long Sandwich(SANDWICH).

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