hdu 4767 Bell
从易做图查到两个公式:易做图网址:http://en.易做图.org/wiki/Bell_number
通过这两个公式可以把Bn分别对于mod拆分成的五个素数取余,然后通过中国剩余定理把这五个素数得到的结果组合起来
对于取余的算法是采用递推的方式
#include <iostream> #include <stack> #include <queue> #include <cstdio> #include <cstdlib> #include <cmath> #include <set> #include <vector> #include <cstring> #include <algorithm> #define INF 0x7ffffffffffLL #define N 50 #define LL long long #define mod 95041567 using namespace std; LL num[N][N], q[N]; int p[5] = {31, 37, 41, 43, 47}, arr[5]; void Preprocessing(){ q[0] = q[1]= 1; num[0][0] = num[1][0] = 1, num[1][1] = 2; for(int i = 2; i < 50; ++i){ q[i] = num[i][0] = num[i - 1][i - 1]; for(int j = 1; j <= i; ++j) num[i][j] = (num[i - 1][j - 1] + num[i][j - 1]) %mod; } } int cal(int x, LL n){ int _q[N]; for(int i = 0; i <= x; ++i) _q[i] = q[i] % x; int len = 0; int _mod[15]; while(n){ _mod[len++] = n % x; n /= x; } int _arr[N]; for(int i = 1; i < len; ++i) for(int j = 1; j <= _mod[i]; ++j){ for(int k = 0; k < x; ++k) _arr[k] = (_q[k] * i + _q[k + 1]) % x; _arr[x] = (_arr[0] + _arr[1]) % x; for(int k = 0; k <= x; ++k) _q[k] = _arr[k]; } return _arr[_mod[0]]; } void 易做图(LL a, LL b, LL &x, LL &y){ if(!b){ x = 1; y = 0; // printf("d = %I64d x = %I64d\n", d, x); return; } 易做图(b, a % b, y, x); y -= a / b * x; } LL china(){ LL d, y, x = 0; for(int i = 0; i < 5; ++i){ LL w = mod / p[i]; 易做图(p[i], w, d, y); x = (x + y * w * arr[i]) % mod; } return (x + mod) % mod; } int main() { // freopen("in.txt","r",stdin); Preprocessing(); int t; scanf("%d", &t); while(t--){ LL n; scanf("%I64d", &n); if(n < 50){ printf("%I64d\n", q[n]); continue; } for(int i = 0; i < 5; ++i) arr[i] = cal(p[i], n); // for(int i = 0; i < 5; ++i) printf("%I64d ", arr[i]); printf("%I64d\n", china()); } return 0; }
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