TC704div2 C ModEquationEasy 矩阵快速幂+dp

Problem Statement

You are given the ints n, K, and v. 



Consider the following modular equation with n variables: (x[0] * x[1] * x[2] * ... * x[n-1]) mod K = v. How many solutions does it have? 



Formally, we want to find the number of sequences (x[0], x[1], ..., x[n-1]) such that each x[i] is an integer between 0 and K-1, inclusive, and the product of all x[i] gives the remainder v when divided by K. 



Please compute and return the number of such sequences, modulo 10^9+7.

Definition

Class:	ModEquationEasy
Method:	count
Parameters:	int, int, int
Returns:	int
Method signature:	int count(int n, int K, int v)
(be sure your method is public)

Limits

Time limit (s):	2.000
Memory limit (MB):	256
Stack limit (MB):	256

Constraints

-

n will be between 1 and 1,000,000,000, inclusive.

-

K will be between 2 and 100, inclusive.

-

v will be between 0 and (K-1), inclusive.

Examples

0)

2

4

1

Returns: 2

The input describes the modular equation (x[0] * x[1]) mod 4 = 1. 



There are two solutions: (1, 1) and (3, 3).

1)

2

4

0

Returns: 8

This time we have 8 solutions: (0, 0), (0, 1), (0, 2), (0, 3), (1, 0), (2, 0), (3, 0) and (2, 2).

2)

2

4

2

Returns: 4

3)

3

7

5

Returns: 36

4)

10

100

50

Returns: 683036071

5)

1

2

0

Returns: 1

题意：

给你n,v,k,找出n个数，使得 (x[0] * x[1] * x[2] * ... * x[n-1]) mod K = v，其中x[i]在0~k之间

思路：

mat[i][j]:=最后一个数取成i，余数为j的个数

转移： mat[i][j] = mat[i][j]+mat[i][k]*mat[k][j];

用K*K的矩阵表示，每乘一次代表取了下一个数。

最后的答案是 mat[i][v] {0<=i<=k}

代码：

int N,K,V;
struct node{
    ll m[105][105];
    void clear(){
        mem(m);
    }
}a;

node mul(node a,node b){
    node re; re.clear();
    for(int i=0; i<K; i++)
        for(int k=0; k<K; k++){
            if(a.m[i][k]){
                for(int j=0; j<K; j++)
                    re.m[i][j] = (re.m[i][j]+(a.m[i][k]*b.m[k][j])%mod)%mod;
            }
        }
    return re;
}

node qpow(int k){
    node re; re.clear();
    for(int i=0; i<=K; i++) re.m[i][i]=1;
    while(k){
        if(k&1) re = mul(re,a);
        a = mul(a,a);
        k >>= 1;
    }
    return re;
}

int count(int n, int k, int v)  
{
    N=n,K=k,V=v;
    a.clear();
     for(int i=0; i<K; i++){
         for(int j=0; j<K; j++)
             a.m[i][i*j%K]++;
     }   
     a = qpow(n-1);
     ll ans = 0;
     for(int i=0; i<K; i++){
         ans = (ans+a.m[i][v])%mod;
     }
     return ans;
}