[CF 582D]Number of Binominal Coefficients

Posted 2020-10-11

题目大意:

给定质数p,正整数α,A,求使\({\alpha ^p}|C_m^n(0 \le m \le n \le A \le {10^{1000}})(m,n \in N)\)成立的(m,n)组数.

 

介绍一个恶心的定理:库默尔定理,定理描述如下:

组合数\(C_m^{m + n}\)所含的p的幂次数为

\(\sum\limits_{i = 1}^\infty  {\left( {\left[ {\frac{{m + n}}{{{p^i}}}} \right] - \left[ {\frac{n}{{{p^i}}}} \right] - \left[ {\frac{m}{{{p^i}}}} \right]} \right)} \)

我们发现上式等同于(m + n)在p进制下进位次数.

 

我们将A转化为p进制,用f[i][j][k][l]表示处理到i位,p的j次方,是否为边,有无进位组合数个数.

 

 1 #include <cstdio>
 2 #include <cstring>
 3 #include <algorithm>
 4 #define N 3600
 5 #define MOD 1000000007
 6 #define __min(a,b) (((a)<(b))?(a):(b))
 7 int A[N],p,a,len,n = 0;char B[N];int f[N][N][2][2];
 8 inline int div(char *a,int p)
 9 {
10     register int r = 0;
11     for(register int i = len;i;i--)
12     {
13         register long long bak = r * 10LL + a[i];
14         r = bak % p;
15         a[i] = bak / p;
16     }
17     for(;len && !a[len];len--);
18     return r;
19 }
20 inline int mul(int a,int b){return a * 1ll * b % MOD;}
21 inline int dec(int a,int b){return (a - b + MOD) % MOD;}
22 inline int gp(int c,int r,int nr){if(r) return dec(p,gp(c,0,nr));return c - nr + 1;}
23 inline int gs(int c,int r,int nr)
24 {
25     if(r) return dec(mul(c + 1,p),gs(c,0,nr));
26     if(c & 1) return dec(mul(c + 2,(c + 1) >> 1),nr * (c + 1));
27     return dec(mul(c + 1,(c + 2) >> 1),nr * (c + 1));
28 }
29 inline void inc(int &a,int b){a = (a + b) % MOD;}
30 int main()
31 {
32     scanf("%d %d",&p,&a);
33     scanf("%s",B + 1);len = strlen(B + 1);
34     for(register int i = 1;i <= len;i++)B[i] -= 48;
35     std::reverse(B + 1,B + len + 1);//reverse(B,len);
36     for(;len;)A[++n] = div(B,p);
37     std::reverse(A + 1,A + n + 1);//reverse(A,n);
38     if(a > n){printf("0\n");return 0;}
39     f[1][0][1][0] = 1;
40     for(int register i = 1;i <= n;i++) for(int register j = 0;j <= a;j++) for(int register eq = 0;eq < 2;eq++) for(int register rem = 0;rem < 2;rem++) if(f[i][j][eq][rem])
41     {
42         int cur = f[i][j][eq][rem];
43         for(int register nr = 0;nr < 2;nr++)
44         {
45             int register nj = __min(j + nr,a);
46             if(!eq) inc(f[i + 1][nj][eq][nr],mul(cur,gs(p - 1,rem,nr)));
47             else
48             {
49                 inc(f[i + 1][nj][eq][nr],mul(cur,gp(A[i],rem,nr)));
50                 if (A[i]) inc(f[i + 1][nj][0][nr],mul(cur,gs(A[i] - 1,rem,nr)));
51             }
52         }
53     }
54     int ans = 0;for(int register eq = 0;eq < 2;eq++) inc(ans,f[n + 1][a][eq][0]);
55     printf("%d\n",ans);
56     return 0;
57 }