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[];
 ll exgcd(ll a,ll b){
     if(!b)return x=,y=,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<)x+=P;return x%P;}
 ll pow(ll a,ll b,ll P){ll t=;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 = , ll b = ): 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[] = ; i < P; i++)
         if(i%B) s[i]=s[i-]*i%P;
         else s[i] = s[i-];
     s[P] = s[P-];
 }
 int main(){
     B = , P = , 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 ; //EOF
     while (c != '-' && (c < '' || c > '') ) c = getchar();
     sgn = (c == '-') ? - : ;
     ret = (c == '-') ?  : (c - '');
     while (c = getchar(), c >= '' && c <= '') ret = ret *  + (c - '');
     ret *= sgn;
     return ;
 }
 template<typename N,typename PN>inline N flo(N a,PN b){return a>=?a/b:-((-a-)/b)-;}
 template<typename N,typename PN>inline N cei(N a,PN b){return a>?(a-)/b+:-(-a/b);}
 template<typename T>inline int sgn(T a) {return a>?:(a<?-:);}
 template<class T> int countbit(const T &n) { return (n==)?:(+countbit(n&(n-))); }
 template <class T1, class T2>
 bool gmax(T1 &a, const T2 &b) { return a < b? a = b, :;}
 template <class T1, class T2>
 bool gmin(T1 &a, const T2 &b) { return a > b? a = b, :;}
 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<>(t) << ", " << get<>(t) << ", " << get<>(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() a = a*(ll)a%;
 }
 const ll inf = 0x3f3f3f3f;
 const ll INF = 1e17;
 const int mod = 1e9+;
 const double eps = 1e-;
 const int N = +;
 const double pi = acos(-1.0);

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

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

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

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

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

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