hdu 3483 矩阵快速幂+二项式定理

Description

This is a very simple problem. Given three integers N, x, and M, your task is to calculate out the following value: 

Input

There are several test cases. For each case, there is a line with three integers N, x, and M, where 1 ≤ N, M ≤ 2*10  ⁹, and 1 ≤ x ≤ 50. 
The input ends up with three negative numbers, which should not be processed as a case. 

Output

For each test case, print a line with an integer indicating the result.

Sample Input

100 1 10000
3 4 1000
-1 -1 -1

Sample Output

5050
444

例：

 

1. Sn=1^x * x^1 + 2^x * x^2 +...+ n^x * x^n; 

2. Sn+1=1^x * x^1 + 2^x * x^2 +...+ n^x * x^n+(n+1)^x * x^(n+1)=Sn+(n+1)^x * x^(n+1),将(n+1)^x二项式展开然后用矩阵快速幂，x^(n+1)则每次用乘x递推

 

 构造矩阵:
 |1 xC(x,0) xC(x,1) xC(x,2) ... xC(x,x)|  |Sn       | |S(n+1)           |
 |0 xC(0,0) 0       0       ... 0      |  |x^n * n^0| |x^(n+1) * (n+1)^0|
 |0 xC(1,0) xC(1,1) 0       ... 0      | *|x^n * n^1|=|x^(n+1) * (n+1)^1|
 |0 xC(2,0) xC(2,1) xC(2,2) ... 0      |  |x^n * n^2| |x^(n+1) * (n+1)^2|
 |...                                  |  |...      | |...              |
 |0 xC(x,0) xC(x,1) xC(x,2) ... xC(x,x)|  |x^n * n^x| |x^(n+1) * (n+1)^x|

#include
#define inf 0x3f3f3f3f
#define ll long long
using namespace std;
struct node
{
    ll mat[60][60];
};
ll n,m;
int x;
ll c[60][60];
void cc()
{
    c[0][0]=c[1][0]=c[1][1]=1;
    for(int i=2;i<=55;i++)
    {
        c[i][0]=c[i][i]=1;
        for(int j=1;j>n>>x>>m)
    {
        if(n==-1)
            return 0;
        node m1;
        memset(m1.mat,0,sizeof(m1.mat));
        m1.mat[0][0]=1;
        for(int i=1;i<=x+1;i++)
            m1.mat[0][i]=(x*c[x][i-1])%m;
        for(int i=1;i<=x+1;i++)
        {
            for(int j=1;j<=i;j++)
                m1.mat[i][j]=(x*c[i-1][j-1])%m;
        }
        node s;
        memset(s.mat,0,sizeof(s.mat));
        for(int i=0;i<=x+1;i++)
            s.mat[i][i]=1;
        int k=n-1;
        while(k)
        {
            if(k%2==1)
                s=mult(s,m1);
            k=k/2;
            m1=mult(m1,m1);
        }
        ll sum=0;
        for(int i=0;i<=x+1;i++)
            sum=(sum+x*s.mat[0][i])%m;
        cout<