【矩阵快速幂】Recurrences UVA - 10870

Think:

1知识点：矩阵快速幂

2题意：


现输入d, n, m求解f(n)

注：f(i) = f(i) mod m;

3题意思考：

（１）：写出系数矩阵，矩阵快速幂求解

4反思：

（１）：注意系数矩阵与初始序列的对应关系

（２）：输入数据中每组测试数据换行相隔，而并不是要求在输出中每组测试数据以换行相隔，注意读题的严谨，注意细节

以下为Accepted代码

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

LL mod;

struct Matrix{
LL v[24][24];
};

Matrix multiply(const Matrix &a, const Matrix &b, int Matrix_len);
Matrix matrix_pow(Matrix x, LL k, int Matrix_len);

LL rec[24];

int main(){
int d, i;
LL n;
while(~scanf("%d %lld %lld", &d, &n, &mod) && (d || n || mod)){
Matrix x;
memset(x.v, 0, sizeof(x.v));
for(i = 1; i <= d; i++){
scanf("%lld", &x.v[1][i]);
x.v[1][i] %= mod;
}
for(i = 1;i < d; i++){
x.v[i+1][i] = 1;
}

for(i = d; i >= 1; i--){/*初始序列：[f(d)...f(1)]*/
scanf("%lld", &rec[i]);
rec[i] %= mod;
}
if(n <= d) printf("%lld\n", rec[d-n+1]%mod);
else {
Matrix y = matrix_pow(x, n-d, d);
LL ans = 0;
for(i = 1; i <= d; i++){
ans += y.v[1][i]*rec[i]%mod;
ans %= mod;
}
printf("%lld\n", ans);
}
}
return 0;
}
Matrix multiply(const Matrix &a, const Matrix &b, int Matrix_len){
Matrix tmp;
memset(tmp.v, 0, sizeof(tmp.v));
for(int i = 1; i <= Matrix_len; i++){
for(int j = 1; j <= Matrix_len; j++){
for(int k = 1; k <= Matrix_len; k++){
tmp.v[i][j] += (a.v[i][k]*b.v[k][j]);
tmp.v[i][j] %= mod;
}
}
}
return tmp;
}
Matrix matrix_pow(Matrix x, LL k, int Matrix_len){
Matrix tmp;
memset(tmp.v, 0, sizeof(tmp.v));
for(int i = 1; i <= Matrix_len; i++)
tmp.v[i][i] = 1;
while(k){
if(k & 1)
tmp = multiply(tmp, x, Matrix_len);
x = multiply(x, x, Matrix_len);
k >>= 1;
}
return tmp;
}