矩阵(mn)基本运算:
+/-:对应位上的数加减;
X:(nn)矩阵 ,c[i][j]=(c[i][j]+a[i][k]*b[k][j]);
对应代码:
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
c.m[i][j]=a.m[i][j]+(-)b.m[i][j];//+/-
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
for(int k=1;k<=n;k++)
c.m[i][j]=c.m[i][j]+a.m[i][k]*b.m[k][j];//*
结构体的定义:
虽可直接定义数组来使用,但为了更加简明,清楚,定义结构体
又因使用C++编译器,无需特地使用typedef struct,所以在此直接使用struct;
struct maxtrix
{
int m[maxn][maxn]
};
int main()
{
maxtrix c;//使用
}
单位矩阵:相当于乘法中的1
int main()
{
maxtrix c;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
c.m[i][j]=(i==j);
return 0;
}
快速幂(数):
二进制的使用
int quickpower(int a,int k)//a^k
{
int c=1;
while(k)
{
if(k&1) c*=a;
k>>=1;
a*=a;
}
return c;
}
快速幂(矩阵):
类似,多了一步矩阵的乘法
maxtrix mul(maxtrix a,maxtrix b)
{
maxtrix c;
memset(c.m,0,sizeof(c.m));
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
for(int k=1;k<=n;k++)
c[i][j]=(c[i][j]+a[i][k]*b[k][j])%mod;//通常数大,需取模
return c;
}
maxtrix quickpower(maxtrix a,int k)//a^k
{
maxtrix c;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
c[i][j]=(i==j);
while(k)
{
if(k&1) c=mul(c,a);
k>>=1;
a=mul(a,a);
}
return c;
}
例题:
1:Tr A//熟悉基本操作
Input
数据的第一行是一个T,表示有T组数据。
每组数据的第一行有n(2 <= n <= 10)和k(2 <= k < 10^9)两个数据。接下来有n行,每行有n个数据,每个数据的范围是[0,9],表示方阵A的内容。
Output
对应每组数据,输出Tr(A^k)%9973。
Sample Input
2
2 2
1 0
0 1
3 99999999
1 2 3
4 5 6
7 8 9
Sample Output
2
2686
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int mod=9973;
int n;
typedef struct
{
int m[15][15];
}maxtrix;//比较sruct maxtrix//构建结构体
maxtrix mul(maxtrix a,maxtrix b)
{
maxtrix c;
memset(c.m,0,sizeof(c.m));
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
for(int k=1;k<=n;k++)
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
return c;
}//矩阵的乘法
maxtrix quickpower(maxtrix a,int k)
{
maxtrix c;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
c.m[i][j]=(i==j);
while(k)
{
if(k&1)
c=mul(c,a);
k/=2;
a=mul(a,a);
}
return c;
}//快速幂操作
int main()
{
int t,k;
maxtrix c;
cin>>t;
while(t--)
{
int sum=0;
cin>>n>>k;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
cin>>c.m[i][j];
c=quickpower(c,k);
for(int i=1;i<=n;i++)
sum=(sum+c.m[i][i])%mod;
cout<<sum<<endl;
}
return 0;
}
2:Fibonacci//基本的应用
In the Fibonacci integer sequence, F0 = 0, F1 = 1, and Fn = Fn − 1 + Fn − 2 for n ≥ 2. For example, the first ten terms of the Fibonacci sequence are:0, 1, 1, 2, 3, 5, 8, 13, 21, 34, …An alternative formula for the Fibonacci sequence is.
Given an integer n, your goal is to compute the last 4 digits of Fn.
Input
The input test file will contain multiple test cases. Each test case consists of a single line containing n (where 0 ≤ n ≤ 1,000,000,000). The end-of-file is denoted by a single line containing the number −1.
Output
For each test case, print the last four digits of Fn. If the last four digits of Fn are all zeros, print ‘0’; otherwise, omit any leading zeros (i.e., print Fn mod 10000).
Sample Input
0
9
999999999
1000000000
-1
Sample Output
0
34
626
6875
Hint
As a reminder, matrix multiplication is associative, and the product of two 2 × 2
matrices is given by.Also, note that raising any 2 × 2 matrix to the 0th power gives the identity matrix:.
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
const int mod=10000;
int n;
typedef struct
{
int m[2][2];
}maxtrix;//构造
maxtrix mul(maxtrix a,maxtrix b)
{
maxtrix c;
memset(c.m,0,sizeof(c.m));
for(int i=0;i<=1;i++)
for(int j=0;j<=1;j++)
for(int k=0;k<=1;k++)
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
return c;
}//矩阵乘法
maxtrix quickpower(maxtrix a)
{
maxtrix c;
c.m[0][0]=c.m[0][1]=c.m[1][0]=1;c.m[1][1]=0;
while(n)
{
if(n&1)
c=mul(c,a);
n/=2;
a=mul(a,a);
}
return c;
}//快速幂
int main()
{
while(cin>>n&&(n!=-1))
{
if(n==0) {cout<<0<<endl;continue; }
maxtrix c;
c.m[0][0]=c.m[0][1]=c.m[1][0]=1;c.m[1][1]=0;
/*for(int i=0;i<=1;i++)
{
for(int j=0;j<=1;j++)
cout<<c.m[i][j]<<" ";
cout<<endl;
} */
c=quickpower(c);
/*for(int i=0;i<=1;i++)
{
for(int j=0;j<=1;j++)
cout<<c.m[i][j]<<" ";
cout<<endl;
} */
cout<<c.m[1][1]%mod<<endl; //
}
return 0;
}
3:Matrix Power Series//二分求和/矩阵分解
看了题解 作者解释得很详细
Given a n × n matrix A and a positive integer k, find the sum S = A + A2 + A3 + … + Ak.
Input
The input contains exactly one test case. The first line of input contains three positive integers n (n ≤ 30), k (k ≤ 109) and m (m < 104). Then follow n lines each containing n nonnegative integers below 32,768, giving A’s elements in row-major order.
Output
Output the elements of S modulo m in the same way as A is given.
Sample Input
2 2 4
0 1
1 1
Sample Output
1 2
2 3
主要注意点S = A + A2 + A3 + … + Ak.;
法一:二分求和
//二分求和
//当k为偶数时:
//s(k)=s(k/2)+A^(n/2) * s(k/2) 即s(k)=(E+A^(n/2))*s(n/2) (E为单位矩阵)
//当k为奇数时:
//s(k)=s(k-1)+A^k , 那么k-1为偶数。依照上面二分
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int n,k,mod;
struct maxtrix
{
int m[35][35];
};//构造
maxtrix add(maxtrix a,maxtrix b)
{
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
a.m[i][j]=(a.m[i][j]+b.m[i][j])%mod;
return a;
}//矩阵加法
maxtrix mul(maxtrix a,maxtrix b)
{
maxtrix c;
memset(c.m,0,sizeof(c.m));
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
for(int k=1;k<=n;k++)
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
return c;
}//矩阵乘法
maxtrix quickpower(maxtrix a,int k1)
{
maxtrix c;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
c.m[i][j]=(i==j);
while(k1)
{
if(k1&1) c=mul(c,a);
k1/=2;
a=mul(a,a);
}
return c;
}//快速幂
maxtrix cal(maxtrix a,int k)
{
if(k==1) return a;
if(k&1)
return add(quickpower(a,k),cal(a,k-1));
k/=2;
return mul(add(quickpower(a,0),quickpower(a,k)),cal(a,k));
}//***二分求和,递归***
int main()
{
cin>>n>>k>>mod;
maxtrix c;
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
cin>>c.m[i][j];
c=cal(c,k);
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
j==n?cout<<c.m[i][j]<<endl:cout<<c.m[i][j]<<" ";
return 0;
}
法二:矩阵分解
//矩阵分解
//|A E|
//|0 E|
//另有一种二分法,
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
int n,k,mod;
typedef struct
{
int m[70][70];
}maxtrix;//构造
maxtrix mul(maxtrix a,maxtrix b)
{
maxtrix c;
memset(c.m,0,sizeof(c.m));
for(int i=1;i<=n;i++)
for(int j=1;j<=n;j++)
for(int k=1;k<=n;k++)
c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
return c;
}//矩阵乘法
int main()
{
cin>>n>>k>>mod;
maxtrix c,b;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
cin>>c.m[i][j];
c.m[i][i+n]=c.m[i+n][i+n]=b.m[i][i]=b.m[i+n][i+n]=1;
} //转换新矩阵
n*=2;k++;
while(k)
{
if(k&1) b=mul(b,c);
/*for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
cout<<b.m[i][j]<<" ";
cout<<endl;
} */
k/=2;
c=mul(c,c);
}
/*for(int i=1;i<=n;i++)
{
for(int j=1;j<=n;j++)
cout<<b.m[i][j]<<" ";
cout<<endl;
} */
n/=2;
for(int i=1;i<=n;i++)
{
b.m[i][i+n]--; //不要忘了减掉
while(b.m[i][i+n]<0)//注意!!由于上一步,可能出现负数,要把mod掉的数加回来
b.m[i][i+n]+=mod;
}
for(int i=1;i<=n;i++)
{
for(int j=1;j<n;j++) cout<<b.m[i][n+j]<<" ";
cout<<b.m[i][2*n]<<endl;
}
return 0;
}