8.3.3 Pendant

Pendant

Time Limit: 6000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 58 Accepted Submission(s): 41

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

思路：此题好题！！！先列出DP方程，再用矩阵加速，网上有很多资料

改天把我的矩阵（照片）发上来～～

#include <cmath>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cstdlib>
using namespace std;

typedef long long ll;
typedef double dd;
const int maxn=35;//change!!
const ll mod=1234567891;
struct matrix
{
    ll m[maxn][maxn];
} c,base,ans,a;
int n,k,T;
ll r;

void close()
{
exit(0);
}

void print(matrix a)
{
    for (int i=1;i<=k+1;i++)
    {
        for (int j=1;j<=k+1;j++)
            printf("%I64d ",a.m[i][j]);
        printf("
");
    }
}

matrix mul(matrix a,matrix b)
{
    memset(c.m,0,sizeof(c.m));
    for (int i=1;i<=k+1;i++)
        for (int j=1;j<=k+1;j++)
            for (int l=1;l<=k+1;l++)
            {
                c.m[i][j]+=(a.m[i][l]*b.m[l][j]);
                c.m[i][j] %= mod;
            }
    return c;
}

void work()
{
    memset(base.m,0,sizeof(base.m));
    memset(ans.m,0,sizeof(ans.m));
    for (int i=1;i<=k+1;i++)
        ans.m[i][i]=1;
    for (int i=1;i<=k;i++)
    {
        base.m[i][i]=i;
        base.m[i][i-1]=k-(i-1);
    }
    base.m[k+1][k+1]=base.m[k+1][k]=1;
    //n++;
    while (n!=0)
    {
        if (n & 1)
        {
            ans=mul(base,ans);
        }
        n/=2;
        base=mul(base,base);
    }
    r=(k * ans.m[k+1][1]) % mod;
    printf("%I64d
",r);
}

void init()
{
    scanf("%d",&T);
    while (T--)
    {
        scanf("%d %d",&n,&k);
        work();
    }
}

int main ()
{
    init();
    close();
    return 0;
}