hihocoder1639 图书馆 [数学]

已知数组a[]及其和sum, 求sum! / (a1!a2!...an!) 的个位数的值。

求某数的逆元表写成了求某数阶乘的逆元表，故一直没找到错误。

P 是质数的幂
B 表示质数,P 表示模数,cal(n) 将返回 n!,以 a × B^b 形式表示，a为模P的情况下。

ll n,x,y,P,B,s[2000000];
ll exgcd(ll a,ll b){
    if(!b)return x=1,y=0,a;
    ll d=exgcd(b,a%b),t=x;
    return x=y,y=t-a/b*y,d;
}
ll rev(ll a,ll P){exgcd(a,P);while(x<0)x+=P;return x%P;}
ll pow(ll a,ll b,ll P){ll t=1;for(;b;b>>=1LL,a=a*a%P)if(b&1LL)t=t*a%P;return t;}
struct Num{
    ll a,b;
    Num(ll a = 1, ll b = 0): a(a), b(b){}
    Num operator*(Num x){return Num(a*x.a%P, b+x.b);}
    Num operator/(Num x){return Num(a*rev(x.a,P)%P, b-x.b);}
};
Num cal(ll n){return n? Num(s[n%P]*pow(s[P],n/P,P)%P,n/B)*cal(n/B): Num();}
void pre(){
    for(int i = s[0] = 1; i < P; i++) 
        if(i%B) s[i]=s[i-1]*i%P; 
        else s[i] = s[i-1];
    s[P] = s[P-1];
}
int main(){
    B = 2, P = 512, pre();
    cal(n);
}

hihocoder1639 别人的题解

自己的题解如下：

#include <bits/stdc++.h>

#define ll long long
#define ull unsigned long long
#define st first
#define nd second
#define pii pair<int, int>
#define pil pair<int, ll>
#define pli pair<ll, int>
#define pll pair<ll, ll>
#define tiii tuple<int, int, int>
#define pw(x) ((1LL)<<(x))
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define sqr(x) ((x)*(x))
#define SIZE(A) ((int)(A.size()))
#define LENGTH(A) ((int)(A.length()))
#define FIN freopen("A.in","r",stdin);
#define FOUT freopen("A.out","w",stdout);
using namespace std;
/***********/
template<typename T>
bool scan (T &ret) {
    char c;
    int sgn;
    if (c = getchar(), c == EOF) return 0; //EOF
    while (c != '-' && (c < '0' || c > '9') ) c = getchar();
    sgn = (c == '-') ? -1 : 1;
    ret = (c == '-') ? 0 : (c - '0');
    while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
    ret *= sgn;
    return 1;
}
template<typename N,typename PN>inline N flo(N a,PN b){return a>=0?a/b:-((-a-1)/b)-1;}
template<typename N,typename PN>inline N cei(N a,PN b){return a>0?(a-1)/b+1:-(-a/b);}
template<typename T>inline int sgn(T a) {return a>0?1:(a<0?-1:0);}
template<class T> int countbit(const T &n) { return (n==0)?0:(1+countbit(n&(n-1))); }
template <class T1, class T2>
bool gmax(T1 &a, const T2 &b) { return a < b? a = b, 1:0;}
template <class T1, class T2>
bool gmin(T1 &a, const T2 &b) { return a > b? a = b, 1:0;}
template <class T> inline T lowbit(T x) {return x&(-x);}

template<class T1, class T2>
ostream& operator <<(ostream &out, pair<T1, T2> p) {
    return out << "(" << p.st << ", " << p.nd << ")";
}
template<class A, class B, class C>
ostream& operator <<(ostream &out, tuple<A, B, C> t) {
    return out << "(" << get<0>(t) << ", " << get<1>(t) << ", " << get<2>(t) << ")";
}
template<class T>
ostream& operator <<(ostream &out, vector<T> vec) {
    out << "("; for(auto &x: vec) out << x << ", "; return out << ")";
}
void testTle(int &a){
    while(1) a = a*(ll)a%1000000007;
}
const ll inf = 0x3f3f3f3f;
const ll INF = 1e17;
const int mod = 1e9+7;
const double eps = 1e-5;
const int N = 100000+10;
const double pi = acos(-1.0);

/***********/


int quick(int x, long long n, int mod) {
    int ans = 1;
    while(n) {
        if(n&1) ans = ans*x%mod;
        x = x*x%mod;
        n >>= 1;
    }
    return ans;
}

long long get2(long long n) {
    long long ans = 0;
    while(n >>= 1)
        ans += n;
    return ans;
}

int m[] = {1, 1, 2, 1, 4}; //阶乘%5
int inv[] = {1, 1, 3, 2, 4}; //i的逆元，写成i!的逆元，狂WA
pair<long long, int> get5(long long n) {
    if(n < 5) return {0, m[n]};
    pair<long long, int> ret = get5(n/5);
    ret.st += n/5;
    ret.nd = ret.nd*quick(m[4], n/5, 5)*m[n%5]%5;
    return ret;
}



int main() {
    int T; scanf("%d", &T);
    long long a[1000];
    while(T--) {
        int n; scanf("%d", &n);
        long long sum = 0;
        for(int i = 0; i < n; i++)
            scanf("%lld", a+i), sum += a[i];
        long long mod2 = get2(sum);
        auto mod5 = get5(sum);
        for(int i = 0; i < n; i++) {
            mod2 -= get2(a[i]);
            auto ret = get5(a[i]);
            mod5.st -= ret.st;
            mod5.nd = mod5.nd*inv[ret.nd]%5;
        }
        int ans;
        if(mod5.st) ans = mod2? 0: 5;
        else {
            ans = mod5.nd;
            if(mod2) {
                if(ans&1) ans = (ans+5)%10;
            }
            else {
                if(!(ans&1)) ans = (ans+5)%10;
            }
        }
        printf("%d
", ans);
    }
    return 0;
}

附：

一句话阐明如何求阶乘的末尾非0数：

求末尾非0数模5的值，n = 5k时，

n! = (1*2*3*4) * (6*7*8*9) * ... * (5k-4)*(5k-3)*(5k-2)*(5k-1)  *5^k * k!

    = (1*2*3*4/2) * (6*7*8*9/2) * ... * [(5k-4)*(5k-3)*(5k-2)*(5k-1)/2]  *10^k * k!

    = (1*2*3*4/2) * (6*7*8*9/2) * ... * [(5k-4)*(5k-3)*(5k-2)*(5k-1)/2]  * k!  (去掉末尾的几个零，结果不变)

    = (1*2*3*4/2) * (6*7*8*9/2) * ... * [(5k-4)*(5k-3)*(5k-2)*(5k-1)/2]  *6^k * k! （乘6，模5下末尾不变）

    = (1*2*3*4*3) * (6*7*8*9*3) * ... * [(5k-4)*(5k-3)*(5k-2)*(5k-1)*3]  * k!

    = 2^k * k!

阶乘末两位非0数？

P = 4, B = 2;

P = 25, B = 5

再合并一下~

组合数求模