矩阵快速幂（模板）

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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.

代码如下：

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;struct Matrix{    int  m[5][5];} I,A,T;const int mod = 10007;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]%=mod;            }        }    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);//注意n-2为负的情况            ans=0;            printf("Case %d: %d\n",flag++,T.m[1][4]%mod);        }        else            printf("Case %d: %d\n",flag++,0);    }    return 0;}

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 <cstdio>#include <cstring>#include <iostream>#include <algorithm>using namespace std;struct Matrix{int mat[2][2];};Matrix A, f;int n,mod;Matrix mul(Matrix a,Matrix b){Matrix 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;}Matrix quickP(int k){Matrix p = A ,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;//初值为2X2的矩阵int k,t;int a,b;//f[0]= a, f[1] = b;int z,w;int cas=0;while(scanf("%d",&t)!=EOF){while(t--){scanf("%d%d%d%d",&a,&b,&z,&w);mod=1;A.mat[1][1]=0;//初始化等比矩阵A.mat[0][0]=A.mat[1][0]=A.mat[0][1]=1;for(int i = 1 ; i <= w ; i++)//保留的位数mod*=10;Matrix ret=quickP(z-1);//z-1 乘的次数,注意z-1为负的情况int ans=(ret.mat[0][0]*b%mod+ret.mat[0][1]*a%mod)%mod;printf("Case %d: %d\n",++cas,ans);}}return 0;}

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