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  • hdu 5446(中国剩余+lucas+按位乘)

    题意:c( n, m)%M    M = P1 * P2 * ......* Pk

    Lucas定理是用来求 c(n,m) mod p,p为素数的值。得出一个存余数数组,在结合中国剩余定理求值

    其中有个地方乘积可能超范围,所以按位乘(数论方面薄弱啊,学习学习)。


    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <vector>
    #include <queue>
    using namespace std;
    typedef long long ll;
    
    
    ll p[15],an[15];
    ll fac[100005],inv[100005];
    
    ll pow_mod(ll a, int n, int mod)
    {
        ll ret = 1;
        while (n)
        {
            if (n&1) ret = ret * a % mod;
            a = a * a % mod;
            n >>= 1;
        }
        return ret;
    }
    
    void ini(int x)
    {
        fac[0] = 1;
        for(int i = 1; i < x; i++) fac[i] = fac[i-1]*i%x;
        inv[x - 1] = pow_mod(fac[x-1],x-2,x);
        for(int i = x - 2; i >= 0; i--)   inv[i] = inv[i+1] * (i+1) % x;
    }
    
    ll c(ll a,ll b,ll p)
    {
        if(a < b || a < 0 || b < 0)
            return 0;
        return fac[a]*inv[b]%p*inv[a-b]%p;
    }
    
    ll lucas(ll a,ll b, int p)
    {
        if( b == 0)
            return 1;
        return lucas(a/p,b/p,p)*c(a%p,b%p,p)%p;
    }
    
    
    ll ex_gcd(ll a, ll b, ll& x, ll& y)
    {
        if (b == 0)
        {
            x = 1;
            y = 0;
            return a;
        }
        ll d = ex_gcd(b, a % b, y, x);
        y -= x * (a / b);
        return d;
    }
    
    ll mul(ll a, ll b, ll mod)
    {
        a = (a % mod + mod) % mod;
        b = (b % mod + mod) % mod;
    
        ll ret = 0;
        while(b)
        {
            if(b&1)
            {
                ret += a;
                if(ret >= mod) ret -= mod;
            }
            b >>= 1;
            a <<= 1;
            if(a >= mod) a -= mod;
        }
        return ret;
    }
    
    
    ll china(ll n,ll* a,ll* b)
    {
        ll M = 1,d,y,x= 0;
        for(int i = 0; i < n; i++)
        {
            M *= b[i];
        }
        for(int i = 0; i < n; i++)
        {
            ll w = M/b[i];
            ex_gcd(b[i],w,d,y);
            x = (x + mul(mul(y, w, M), a[i], M));//可能超范围
        }
        return (x+M) % M;
    }
    
    
    int main()
    {
        int T,k;
        ll n,m;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%I64d%I64d",&n,&m);
            scanf("%d",&k);
            for(int i = 0; i < k; i++)
            {
                scanf("%I64d",&p[i]);
                ini(p[i]);
                an[i] = lucas(n,m,p[i]);
            }
            printf("%I64d
    ",china(k,an,p));
        }
        return 0;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/Przz/p/5409757.html
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