Pendant

Time Limit: 6000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 1032    Accepted Submission(s): 535

Problem Description

On
Saint Valentine‘s Day, Alex imagined to present a special pendant to
his girl friend made by K kind of pearls. The pendant is actually a
string of pearls, and its length is defined as the number of pearls in
it. As is known to all, Alex is very rich, and he has N pearls of each
kind. Pendant can be told apart according to permutation of its pearls.
Now he wants to know how many kind of pendant can he made, with length
between 1 and N. Of course, to show his wealth, every kind of pendant
must be made of K pearls.
Output the answer taken modulo 1234567891.

Input

The
input consists of multiple test cases. The first line contains an
integer T indicating the number of test cases. Each case is on one line,
consisting of two integers N and K, separated by one space.
Technical Specification

1 ≤ T ≤ 10
1 ≤ N ≤ 1,000,000,000
1 ≤ K ≤ 30

Output

Output the answer on one line for each test case.

Sample Input

2
2 1
3 2

Sample Output

2
8

Source

The 4th Baidu Cup final

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题解：

设f[i][j] 表示长度为i，用了j种珍珠的方案个数；

我们考虑加一个位置，我们可以让它是之前出现过的珍珠，也可以是没出现过的珍珠；

f[i][j] = (k-(i-1))*f[i-1][j-1] + j*f[i-1][j]；

我们发现这个dp是O(nk)的，n变态的大显然炸掉；

看到n自然而然的会想到矩阵加速；

我们设一个转移矩阵是G，G[k+1][k+1]， 为什么是k+1？

我们要算总的方案个数，要把所有的f[i][k]加起来，所以我们多开一维，用来转移f[i][k]的和；

G的矩阵长这样

1 0 0 0 0 1　　　　　　sum　　　　sum‘　　

0 1 0 0 0 1　　　　　　f1　　　　 f1‘

0 k-1 2 0 0 0    *     f2  -> 　　f2‘

...　　　　　　　　　　...　　　　...

0 0 0 0 1 k　　　　　　fk　　　　 fk‘

就这样

Code：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
#define ll long long
#define mod 1234567891
#define N 32
int T;
ll n, k;

struct Mat
{
    ll a[N][N];
    Mat() {memset(a, 0, sizeof a);}
    inline void clear() {memset(a, 0, sizeof a);}
    inline void ini() {for(int i=0;i<=k;i++)a[i][i]=1;}
    friend Mat operator * (Mat x, Mat y)
    {
        Mat z;
        for (register int p = 0 ; p <= k ; p ++)
        {
            for (register int i = 0 ; i <= k ; i ++)
            {
                for (register int j = 0 ; j <= k ; j ++)
                {
                    z.a[i][j] = (z.a[i][j] + x.a[i][p] * y.a[p][j]) % mod;
                }
            }
        }
        return z;
    }
    friend Mat operator ^ (Mat x, ll y)
    {
        Mat z;z.ini();
        while (y)
        {
            if (y & 1) z = z * x;
            x = x * x;
            y >>= 1;
        }
        return z;
    }
}G, B, C;
inline void init() {G.clear(), B.clear(), C.clear();}

int main()
{
    scanf("%d", &T);
    while (T--)
    {
        init();
        scanf("%lld%lld", &n, &k);
        G.a[0][0] = 1, G.a[0][k] = 1;
        G.a[1][1] = 1;
        for (register int i = 2 ; i <= k ; i ++)
        {
            G.a[i][i] = i;
            G.a[i][i-1] = k - i + 1;
        }
    //    for (int i=0;i<=k;i++,puts(""))for(int j=0;j<=k;j++) printf("%d ",G.a[i][j]) ;
        B.a[1][0] = k;
        C = G ^ n;
        C = C * B;
        cout<<C.a[0][0]<<endl;
    }
    return 0;
}

原文地址：https://www.cnblogs.com/zZh-Brim/p/9261106.html