hdoj 4767 Bell 【矩阵快速幂 + CRT】
BellTime Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 666 Accepted Submission(s): 286
Problem Description
What? MMM is learning Combinatorics!?
Looks like she is playing with the bell sequence now: bell[n] = number of ways to partition of the set {1, 2, 3, ..., n} e.g. n = 3: {1, 2, 3} {1} {2 3} {1, 2} {3} {1, 3} {2} {1} {2} {3} so, bell[3] = 5 MMM wonders how to compute the bell sequence efficiently.
Input
T -- number of cases
for each case: n (1 <= n < 2^31)
Output
for each case:
bell[n] % 95041567
Sample Input
Sample Output
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Bell数：讲解特详细
看会上面这个，就可以KO这道题了。
先找到95041567的小于100的质因子：
#include <cstdio>
int main()
{
int n = 95041567;
while(n != 1)
{
for(int i = 2; i <= 100; i++)//没有必要判断是否是质数 就可以找到。。。
{
if(n % i == 0)
{
printf("%d\n", i);
n /= i;
break;
}
}
}
return 0;
}
我构建的矩阵如下：
AC代码：
#include <cstdio>
#include <cstring>
#include <algorithm>
#define LL long long
#define debug printf("1\n");
using namespace std;
int m[6] = {0, 31, 37, 41, 43, 47};
int rm[6];
int a[6];//方程组x % m = a
int B[6][60][60];//保存对m[i]取余后的 Bell数
int F[60];
void getBell(int id, int MOD)//预处理贝尔数
{
B[id][0][1] = 1;
for(int i = 1; i <= 50; i++)
{
for(int j = 1; j <= i+1; j++)
{
if(j == 1)
B[id][i][j] = B[id][i-1][i];
else
B[id][i][j] = (B[id][i-1][j-1] + B[id][i][j-1]) % MOD;
}
}
}
struct MATRIX
{
struct Matrix{
int a[60][60];
int r, c;
};
Matrix ori, res;
void init(int id, int p)//建立p*p矩阵
{
for(int i = 1; i <= p; i++)
F[i] = B[id][i][1] % p;
//debug;
ori.r = res.r = ori.c = res.c = p;
memset(ori.a, 0, sizeof(ori.a));
memset(res.a, 0, sizeof(res.a));
for(int i = 1; i <= p; i++)
{
res.a[i][i] = 1;
if(i < p)
ori.a[i][i] = ori.a[i+1][i] = 1;
else
ori.a[i][i] = ori.a[1][i] = ori.a[2][i] = 1;
}
}
Matrix multi(Matrix x, Matrix y, int MOD)
{
Matrix z;
memset(z.a, 0, sizeof(z.a));
z.r = x.r; z.c = y.c;
for(int i = 1; i <= x.r; i++)
{
for(int k = 1; k <= x.c; k++)
{
if(x.a[i][k] == 0) continue;
for(int j = 1; j <= y.c; j++)
z.a[i][j] = (z.a[i][j] + (x.a[i][k] * y.a[k][j]) % MOD) % MOD;
}
}
return z;
}
void solve(int n, int MOD)
{
while(n)
{
if(n & 1)
res = multi(ori, res, MOD);
ori = multi(ori, ori, MOD);
n >>= 1;
}
}
};
MATRIX MA;
struct CRT
{
int gcd(int a, int b){
return b == 0 ? a : gcd(b, a%b);
}
void exgcd(int a, int b, int &d, int &x, int &y)
{
if(b == 0) {d = a, x = 1, y = 0;}
else
{
exgcd(b, a%b, d, y, x);
y -= x * (a / b);
}
}
int solve(int l, int r, int *m, int *a)
{
int LCM = 1;
for(int i = l; i <= r; i++)
LCM = LCM / gcd(LCM, m[i]) * m[i];
for(int i = l+1; i <= r; i++)
{
int A = m[l], B = m[i], d, x, y, c = a[i] - a[l];
exgcd(A, B, d, x, y);
if(c % d) return -1;
int mod = m[i] / d;
int k = ((x * c / d) % mod + mod) % mod;
a[l] = m[l] * k + a[l];
m[l] = m[l] / d * m[i];
}
if(a[l] == 0) return LCM;
return a[l];
}
};
CRT crt;
int main()
{
for(int i = 1; i <= 5; i++)
getBell(i, m[i]);
int t;
int n;
scanf("%d", &t);
while(t--)
{
scanf("%d", &n);
if(n <= 50)
{
for(int i = 1; i <= 5; i++)
a[i] = B[i][n][1];
memcpy(rm, m, sizeof(m));
printf("%d\n", crt.solve(1, 5, rm, a));
continue;
}
for(int i = 1; i <= 5; i++)
{
MA.init(i, m[i]);//构建矩阵
int temp = n / m[i];
if(n % m[i] == 0)
temp -= 1;
MA.solve(temp, m[i]);
int remain = n % m[i];//找到位置 即第多少列
if(remain == 0)
remain = m[i];
a[i] = 0;
for(int j = 1; j <= m[i]; j++)//直接求解当前列的值就可以了
a[i] = (a[i] + (MA.res.a[j][remain] * F[j]) % m[i]) % m[i];
}
memcpy(rm, m, sizeof(m));//一开始忘了这个了。。。
printf("%d\n", crt.solve(1, 5, rm, a));//求解
}
return 0;
}