矩阵快速幂模板篇
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或许你们看不太懂，纯属自用；
第一种：
Description
Let‘s define another number sequence, given by the following function:
f(0) = a
f(1) = b
f(n) = f(n-1) + f(n-2), n > 1
When a = 0 and b = 1, this sequence gives the Fibonacci sequence. Changing the values of a and b, you can get many different sequences. Given the values of a, b, you have to find the last m digits of f(n).
Input
Input starts with an integer T (≤ 10000), denoting the number of test cases.
Each test case consists of a single line containing four integers a b n m. The values of a and b range in [0,100], value of n ranges in [0, 109] and value of m ranges in [1, 4].
Output
For each case, print the case number and the last m digits of f(n). However, do NOT print any leading zero.
Sample Input
4
0 1 11 3
0 1 42 4
0 1 22 4
0 1 21 4
Sample Output
Case 1: 89
Case 2: 4296
Case 3: 7711
Case 4: 946
代码如下：
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;
struct A
{
int mat[2][2];
};
A d,f;
int n,mod;
A mul(A a,A b)
{
A t;
memset(t.mat,0,sizeof(t.mat));
for(int i=0;i<n;i++)
{
for(int k=0;k<n;k++)
{
if(a.mat[i][k])
for(int j=0;j<n;j++)
{
t.mat[i][j]+=a.mat[i][k]*b.mat[k][j];
t.mat[i][j]%=mod;
}
}
}
return t;
}
A quickP(int k)
{
A p = d ,m;
memset(m.mat,0,sizeof(m.mat));
for(int i=0;i<n;++i)//单位矩阵
{
m.mat[i][i]=1;
}
/*if(k==0)
return e;*/
while(k)
{
if(k & 1)
m=mul(m,p);
p=mul(p,p);
k >>= 1 ;
}
return m;
}
int main()
{
n=2;
int k,t;int x,y,z,w;
while(scanf("%d",&t)!=EOF)
{
int s=0;
while(t--)
{
scanf("%d%d%d%d",&x,&y,&z,&w);
mod=1;
d.mat[1][1]=0;
d.mat[0][0]=d.mat[1][0]=d.mat[0][1]=1;
for(int i = 1 ; i <= w ; i++)
mod*=10;
A ret=quickP(z-1);//z-1 乘的次数
int ans=(ret.mat[0][0]*y%mod+ret.mat[0][1]*x%mod)%mod;
printf("Case %d: %d\n",++s,ans);
}
}
return 0;
}
Description
You have to find the nth term of the following function:
f(n) = a * f(n-1) + b * f(n-3) + c, if(n > 2)
= 0, if(n ≤ 2)
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case contains four integers n (0 ≤ n ≤ 108),a b c (1 ≤ a, b, c ≤ 10000).
Output
For each case, print the case number and f(n) modulo 10007.
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct Matrix
{
int m[5][5];
}I,A,T;
int a,b,n,mod;
Matrix matrixmul(Matrix a,Matrix b)
{
int i,j,k;
Matrix c;
for (i=1;i<=4;i++)
for(j=1;j<=4;j++)
{
c.m[i][j]=0;
for(k=1;k<=4;k++)
{
c.m[i][j]+=(a.m[i][k]*b.m[k][j]);
c.m[i][j]%=10007;
}
}
return c;
}
Matrix quickpagow(int n)
{
Matrix m=A,b=I;
while (n>=1)
{
if(n&1)
b=matrixmul(b,m);
n=n>>1;
m=matrixmul(m,m);
}
return b;
}
int main()
{
int t,ans,i,flag=1,a,b,c;
scanf("%d",&t);
while(t--)
{
scanf("%d%d%d%d",&n,&a,&b,&c);
memset(I.m,0,sizeof(I.m));
memset(A.m,0,sizeof(A.m));
for(i=1;i<=4;i++)
{
I.m[i][i]=1;
}
A.m[1][1]=a,A.m[1][2]=0,A.m[1][3]=b,A.m[1][4]=c;
A.m[2][1]=A.m[3][2]=A.m[4][4]=1;
if(n>=3)
{
T=quickpagow(n-2);
ans=0;
printf("Case %d: %d\n",flag++,T.m[1][4]%10007);
}
else
printf("Case %d: %d\n",flag++,0);
}
return 0;
}