标签：矩阵快速幂
题目大意：
If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .
现在给你k和m，求f(k) % m。
解题思路：
f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10)
其中k为10^9数量级，必然不能用递推的方式做。这类题目可以通过构造矩阵，用矩阵快速幂来做。
构造的矩阵是:
|0
|0
|................1|
|a9
代码：
/*
ID: wuqi9395@126.com
PROG:
LANG: C++
*/
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<fstream>
#include<cstring>
#include<ctype.h>
#include<iostream>
#include<algorithm>
#define INF (1<<30)
#define PI acos(-1.0)
#define mem(a, b) memset(a, b, sizeof(a))
#define For(i, n) for (int i = 0; i < n; i++)
typedef long long ll;
using namespace std;
const int maxn = 20;
const int maxm = 20;
ll mod, k;
struct Matrix {
int n, m;
int a[maxn][maxm];
void clear() {
n = m = 0;
memset(a, 0, sizeof(a));
}
Matrix operator * (const Matrix &b) const { //实现矩阵乘法
Matrix tmp;
tmp.clear();
tmp.n = n; tmp.m = b.m;
for (int i = 0; i < n; i++)
for (int j = 0; j < b.m; j++)
for (int k = 0; k < m; k++) {
tmp.a[i][j] += a[i][k] * b.a[k][j];
tmp.a[i][j] %= mod;
}
return tmp;
}
};
Matrix A, B;
void init() {
A.clear(); //矩阵A是构造的矩阵
A.n = A.m = 10;
for (int i = 0; i < 9; i++)
A.a[i][i + 1] = 1;
B.clear();
B.n = 10; B.m = 1; //矩阵B是f(x)的前10个数
for (int i = 0; i < 10; i++)
B.a[i][0] = i;
}
Matrix Matrix_pow(Matrix A, ll k, ll mod) { //矩阵快速幂
Matrix res;
res.clear();
res.n = res.m = 10;
for (int i = 0; i < 10; i++) res.a[i][i] = 1;
while(k) {
if (k & 1) res = A * res;
k >>= 1;
A = A * A;
}
return res;
}
int main () {
init();
while(cin>>k>>mod) {
int x;
for (int i = 0; i < 10; i++) {
scanf("%d", &x);
A.a[9][9 - i] = x;
}
if (k < 10) {
printf("%lld\n", k % mod);
}
else {
Matrix res = Matrix_pow(A, k - 9, mod);
res = res * B;
cout<<res.a[9][0]<<endl;
}
}
return 0;
}
[HDU 1757] A Simple Math Problem （矩阵快速幂）,布布扣,bubuko.com
[HDU 1757] A Simple Math Problem （矩阵快速幂）
标签：矩阵快速幂
原文地址：http://blog.csdn.net/sio__five/article/details/37072641
