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。

 1 ll n,x,y,P,B,s[2000000];
 2 ll exgcd(ll a,ll b){
 3     if(!b)return x=1,y=0,a;
 4     ll d=exgcd(b,a%b),t=x;
 5     return x=y,y=t-a/b*y,d;
 6 }
 7 ll rev(ll a,ll P){exgcd(a,P);while(x<0)x+=P;return x%P;}
 8 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;}
 9 struct Num{
10     ll a,b;
11     Num(ll a = 1, ll b = 0): a(a), b(b){}
12     Num operator*(Num x){return Num(a*x.a%P, b+x.b);}
13     Num operator/(Num x){return Num(a*rev(x.a,P)%P, b-x.b);}
14 };
15 Num cal(ll n){return n? Num(s[n%P]*pow(s[P],n/P,P)%P,n/B)*cal(n/B): Num();}
16 void pre(){
17     for(int i = s[0] = 1; i < P; i++) 
18         if(i%B) s[i]=s[i-1]*i%P; 
19         else s[i] = s[i-1];
20     s[P] = s[P-1];
21 }
22 int main(){
23     B = 2, P = 512, pre();
24     cal(n);
25 }
View Code

 

hihocoder1639 别人的题解

自己的题解如下：

  1 #include <bits/stdc++.h>
  2 
  3 #define ll long long
  4 #define ull unsigned long long
  5 #define st first
  6 #define nd second
  7 #define pii pair<int, int>
  8 #define pil pair<int, ll>
  9 #define pli pair<ll, int>
 10 #define pll pair<ll, ll>
 11 #define tiii tuple<int, int, int>
 12 #define pw(x) ((1LL)<<(x))
 13 #define lson l, m, rt<<1
 14 #define rson m+1, r, rt<<1|1
 15 #define sqr(x) ((x)*(x))
 16 #define SIZE(A) ((int)(A.size()))
 17 #define LENGTH(A) ((int)(A.length()))
 18 #define FIN freopen("A.in","r",stdin);
 19 #define FOUT freopen("A.out","w",stdout);
 20 using namespace std;
 21 /***********/
 22 template<typename T>
 23 bool scan (T &ret) {
 24     char c;
 25     int sgn;
 26     if (c = getchar(), c == EOF) return 0; //EOF
 27     while (c != '-' && (c < '0' || c > '9') ) c = getchar();
 28     sgn = (c == '-') ? -1 : 1;
 29     ret = (c == '-') ? 0 : (c - '0');
 30     while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
 31     ret *= sgn;
 32     return 1;
 33 }
 34 template<typename N,typename PN>inline N flo(N a,PN b){return a>=0?a/b:-((-a-1)/b)-1;}
 35 template<typename N,typename PN>inline N cei(N a,PN b){return a>0?(a-1)/b+1:-(-a/b);}
 36 template<typename T>inline int sgn(T a) {return a>0?1:(a<0?-1:0);}
 37 template<class T> int countbit(const T &n) { return (n==0)?0:(1+countbit(n&(n-1))); }
 38 template <class T1, class T2>
 39 bool gmax(T1 &a, const T2 &b) { return a < b? a = b, 1:0;}
 40 template <class T1, class T2>
 41 bool gmin(T1 &a, const T2 &b) { return a > b? a = b, 1:0;}
 42 template <class T> inline T lowbit(T x) {return x&(-x);}
 43 
 44 template<class T1, class T2>
 45 ostream& operator <<(ostream &out, pair<T1, T2> p) {
 46     return out << "(" << p.st << ", " << p.nd << ")";
 47 }
 48 template<class A, class B, class C>
 49 ostream& operator <<(ostream &out, tuple<A, B, C> t) {
 50     return out << "(" << get<0>(t) << ", " << get<1>(t) << ", " << get<2>(t) << ")";
 51 }
 52 template<class T>
 53 ostream& operator <<(ostream &out, vector<T> vec) {
 54     out << "("; for(auto &x: vec) out << x << ", "; return out << ")";
 55 }
 56 void testTle(int &a){
 57     while(1) a = a*(ll)a%1000000007;
 58 }
 59 const ll inf = 0x3f3f3f3f;
 60 const ll INF = 1e17;
 61 const int mod = 1e9+7;
 62 const double eps = 1e-5;
 63 const int N = 100000+10;
 64 const double pi = acos(-1.0);
 65 
 66 /***********/
 67 
 68 
 69 int quick(int x, long long n, int mod) {
 70     int ans = 1;
 71     while(n) {
 72         if(n&1) ans = ans*x%mod;
 73         x = x*x%mod;
 74         n >>= 1;
 75     }
 76     return ans;
 77 }
 78 
 79 long long get2(long long n) {
 80     long long ans = 0;
 81     while(n >>= 1)
 82         ans += n;
 83     return ans;
 84 }
 85 
 86 int m[] = {1, 1, 2, 1, 4}; //阶乘%5
 87 int inv[] = {1, 1, 3, 2, 4}; //i的逆元，写成i!的逆元，狂WA
 88 pair<long long, int> get5(long long n) {
 89     if(n < 5) return {0, m[n]};
 90     pair<long long, int> ret = get5(n/5);
 91     ret.st += n/5;
 92     ret.nd = ret.nd*quick(m[4], n/5, 5)*m[n%5]%5;
 93     return ret;
 94 }
 95 
 96 
 97 
 98 int main() {
 99     int T; scanf("%d", &T);
100     long long a[1000];
101     while(T--) {
102         int n; scanf("%d", &n);
103         long long sum = 0;
104         for(int i = 0; i < n; i++)
105             scanf("%lld", a+i), sum += a[i];
106         long long mod2 = get2(sum);
107         auto mod5 = get5(sum);
108         for(int i = 0; i < n; i++) {
109             mod2 -= get2(a[i]);
110             auto ret = get5(a[i]);
111             mod5.st -= ret.st;
112             mod5.nd = mod5.nd*inv[ret.nd]%5;
113         }
114         int ans;
115         if(mod5.st) ans = mod2? 0: 5;
116         else {
117             ans = mod5.nd;
118             if(mod2) {
119                 if(ans&1) ans = (ans+5)%10;
120             }
121             else {
122                 if(!(ans&1)) ans = (ans+5)%10;
123             }
124         }
125         printf("%d\n", ans);
126     }
127     return 0;
128 }
View Code

 

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

posted @ 2018-04-07 23:43  我在地狱  阅读(187)  评论(0)  编辑  收藏  举报