wenbao与矩阵

矩阵满足分配率和结合律（非常重要）

****但是不满足交换律

循环矩阵：

　　两个循环矩阵乘积依然是循环矩阵

矩阵乘法

struct mat{
    int n, m;
    double data[MAXN][MAXN];
};

int mul(mat& c, const mat& a, const mat& b){
    int i, j, k;
    if (a.m != b.n)
        return 0;
    c.n = a.n;
    c.m = b.m;
    for (i = 0; i < c.n; i++)
        for (j = 0; j < c.m; j++)
            for (c.data[i][j] = k = 0; k < a.m; k++)
                c.data[i][j] += a.data[i][k] * b.data[k][j];
    return 1;
}

矩阵快速幂

struct Node{
    ll x[16][16];
};
Node mul(Node xx, Node yy){
    Node X;
    for(int i = 0; i < d; ++i){
        for(int j = 0; j < d; ++j){
            X.x[i][j] = 0;
            for(int k = 0; k < d; ++k){
                X.x[i][j] = (X.x[i][j] + (xx.x[i][k]*yy.x[k][j])%m) % m;
            }
        }
    }
    return X;
}
Node q_m(Node A, ll x){
    Node AAA;
    for(int i = 0; i < d; ++i){
        for(int j = 0; j < d; ++j){
            AAA.x[i][j] = (i == j);
            //printf("%lld ", AAA.x[i][j]);
        }
        //puts("");
    }
    while(x){
        if(x&1) AAA = mul(AAA, A);
        A = mul(A, A);
        x >>= 1;
    }
    return AAA;
}

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矩阵快速幂入门题一：

http://acm.hdu.edu.cn/showproblem.php?pid=1575

直接求矩阵快速幂

#include <iostream>
using namespace std;
#define ll long long
const ll Mod = 9973;
int n;
ll k;
struct Node{
    ll x[11][11];
}A;
Node mul(Node a, Node b){
    Node AAA;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            AAA.x[i][j] = 0;
            for(int k = 0; k < n; ++k){
                AAA.x[i][j] = (AAA.x[i][j] + a.x[i][k]*b.x[k][j]%Mod)%Mod;
            }
        }
    }
    return AAA;
}
ll q_m(){
    Node AA;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            AA.x[i][j] = (i == j);
        }
    }
    while(k){
        if(k&1) AA = mul(AA, A);
        A = mul(A, A);
        k >>= 1;
    }
    ll sum = 0;
    for(int i = 0; i < n; ++i){
        sum = (sum+AA.x[i][i])%Mod;
    }
    return sum;
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d%lld", &n, &k);
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                scanf("%lld", &A.x[i][j]);
            }
        }
        cout<<q_m()<<endl;
    }
    return 0;
}

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矩阵快速幂入门二：

http://codeforces.com/contest/185/problem/A

递推上三角形的个数。。。。。。。。。

其实直接推也可以的

#include <iostream>
using namespace std;
#define ll long long 
const ll Mod = 1e9+7;
ll n;
struct Node{
    ll x[2][2];
};
Node mul(Node a, Node b){
    Node AAA;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            AAA.x[i][j] = 0;
            for(int k = 0; k < 2; ++k){
                AAA.x[i][j] = (AAA.x[i][j] + a.x[i][k]*b.x[k][j]%Mod)%Mod;
            }
        }
    }
    return AAA;
}
ll q_m(ll x){
    Node AA, A;
    A.x[0][0] = A.x[1][1] = 3LL;
    A.x[0][1] = A.x[1][0] = 1LL;
    AA.x[0][0] = AA.x[1][1] = 1LL;
    AA.x[0][1] = AA.x[1][0] = 0LL;
    while(x){
        if(x&1) AA = mul(AA, A);
        A = mul(A, A);
        x >>= 1;
    }
    return AA.x[0][0];
}
int main(){
    cin >> n;
    cout<<q_m(n)<<endl;
    return 0;
}

#include <iostream>
using namespace std;
#define ll long long 
const ll Mod = 1e9+7;
ll n;
ll q_m(ll x){
    ll xx = 1LL, xxx = 2LL;
    while(x){
        if(x&1) xx = xx*xxx%Mod;
        xxx = xxx*xxx%Mod;
        x >>= 1;
    }
    return xx;
}
int main(){
    cin >> n;
    if(n == 0){
        cout<<1<<endl;
        return 0;
    }
    cout<<(q_m(n-1LL)+q_m(2LL*n-1LL))%Mod<<endl;
    return 0;
}

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矩阵快速幂入门三：

http://acm.hdu.edu.cn/showproblem.php?pid=2157

矩阵基本知识关于通路

#include <iostream>
#include <string.h>
using namespace std;
const int Mod = 1000;
int n, m;
struct Node{
    int x[22][22];
}A;
Node mul(Node a, Node b){
    Node AAA;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            AAA.x[i][j] = 0;
            for(int k = 0; k < n; ++k){
                AAA.x[i][j] = (AAA.x[i][j] + a.x[i][k]*b.x[k][j]%Mod)%Mod;
            }
        }
    }
    return AAA;
}
void q_m(Node A, int xx, int yy, int zz){
    Node AA;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            AA.x[i][j] = (i == j);
        }
    }
    while(zz){
        if(zz&1) AA = mul(AA, A);
        A = mul(A, A);
        zz >>= 1;
    }
    /*
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            printf("%d ", AA.x[i][j]);
        }
        cout<<endl;
    }
    */
    printf("%d
", AA.x[xx][yy]);
}
int main(){
    while(scanf("%d%d", &n, &m) == 2 && n+m){
        memset(A.x, 0, sizeof(A.x));
        while(m--){
            int xx, yy;
            scanf("%d%d", &xx, &yy);
            A.x[xx][yy] = 1;
        }
        int t;
        scanf("%d", &t);
        while(t--){
            int xx, yy, zz;
            scanf("%d%d%d", &xx, &yy, &zz);
            q_m(A, xx, yy, zz);
        }
    }
    return 0;
}

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矩阵快速幂入门四

http://poj.org/problem?id=3070

求斐波那契数列

#include <iostream>
#include <stdio.h>
using namespace std;
#define ll long long
const int Mod = 10000;
struct Node{
    ll x[2][2];
};
Node mul(Node a, Node b){
    Node AAA;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            AAA.x[i][j] = 0;
            for(int k = 0; k < 2; ++k){
                AAA.x[i][j] = (AAA.x[i][j] + a.x[i][k]*b.x[k][j]%Mod)%Mod;
            }
        }
    }
    return AAA;
}
void q_m(ll x){
    Node AA, A;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            AA.x[i][j] = (i == j);
        }
    }
    A.x[0][0] = A.x[0][1] = A.x[1][0] = 1LL, A.x[1][1] = 0;
    while(x){
        if(x&1) AA = mul(AA, A);
        A = mul(A, A);
        x >>= 1;
    }
    /*
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            printf("%lld ", AA.x[i][j]);
        }
        cout<<endl;
    }
    */
    printf("%lld
", AA.x[0][1]);
}
int main(){
    int n;
    while(scanf("%lld", &n) && n != -1){
        q_m(n);
    }
    return 0;
}

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大牛博客（关于递推构造矩阵）强烈推荐

 http://www.cnblogs.com/frog112111/archive/2013/05/19/3087648.html

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https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=20&problem=1811&mosmsg=Submission+received+with+ID+18912945

　　引用博客　　http://blog.csdn.net/u010304217/article/details/39212049

 f[n]=a1*f[n-1]+a2*f[n-2]+...+ad*f[n-d];
    由于n太大，不能直接递推，需要用矩阵快速幂来解决，时间复杂度为O(d^3logn)
    举例，d=5的矩阵关系式为：
                |a1 a2 a3 a4 a5|                 | f[n]     |      | f[n+1] |  
                |1                      |                 | f[n-1]  |      | f[n]    |  
                |     1                 |  *              | f[n-2]  | =   | f[n-1] | (空白处为0)
                |          1            |                 | f[n-3]  |      | f[n-2] |
                |              1        |                 | f[n-4]  |      | f[n-3] |

矩阵快速幂 

#include <iostream>
#include <string.h>
using namespace std;
#define ll long long
ll d, n, m, a[16], f[16];
struct Node{
    ll x[16][16];
};
Node mul(Node xx, Node yy){
    Node X;
    for(int i = 0; i < d; ++i){
        for(int j = 0; j < d; ++j){
            X.x[i][j] = 0;
            for(int k = 0; k < d; ++k){
                X.x[i][j] = (X.x[i][j] + (xx.x[i][k]*yy.x[k][j])%m) % m;
            }
        }
    }
    return X;
}
Node q_m(Node A, ll x){
    Node AAA;
    for(int i = 0; i < d; ++i){
        for(int j = 0; j < d; ++j){
            AAA.x[i][j] = (i == j);
            //printf("%lld ", AAA.x[i][j]);
        }
        //puts("");
    }
    while(x){
        if(x&1) AAA = mul(AAA, A);
        A = mul(A, A);
        x >>= 1;
    }
    return AAA;
}
int main(){
    while(scanf("%lld%lld%lld", &d, &n, &m) && d+n+m){
        ll xx;
        for(int i = 0; i < d; ++i) scanf("%lld", &xx), a[i] = xx%m;
        for(int i = 0; i < d; ++i) scanf("%lld", &xx), f[i] = xx%m;
        if(n <= d){
            printf("%lld
", f[n-1]);
            continue;
        }
        Node A;
        for(int i = 0; i < d; ++i){
            for(int j = 0; j < d; ++j){
                if(i == 0) A.x[i][j] = a[j];
                else A.x[i][j] = (i-1 == j);
                //printf("%lld ", A.x[i][j]);
            }
            //puts("");
        }
        Node AA = q_m(A, n-d);
        ll sum = 0;
        for(int i = 0; i < d; ++i){
            sum = (sum + (AA.x[0][i]*f[d-i -1])%m) % m;
        }
        printf("%lld
", sum);
    }
    return 0;
}

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https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=20&problem=1705&mosmsg=Submission+received+with+ID+2137361

细胞自动机 

循环矩阵性质：循环矩阵相乘依然是循环矩阵，且该循环矩阵是对称的。。。。。。。。。

所以可以将　　n^3log(k)　　降为　　n^2long(k)　　因为只要算一行就可以了

做完这个题有个心得，，就是可以将矩阵的下标作为参数传来传去效率高，空间消耗小。。。。。。。牛X

#include <iostream>
#include <stdio.h>
using namespace std;
#define ll long long
ll m[200][550], a[550];
int num, n, d, mod, k;
int mul(int x, int y){
    num++;
    for(int i = 0; i < n; ++i){
        ll sum = 0;
        for(int j = 0; j < n; ++j){
            sum = (sum + m[x][j] * m[y][(n+j-i)%n] % mod) % mod;
        }
        m[num][i] = sum;
    }
    return num;
}
int q_m(int x){
    int a = 0, b = 1;
    while(x){
        if(x&1) a = mul(a, b);
        b = mul(b, b);
        x >>= 1;
    }
    return a;
}
int main(){
    while(scanf("%d%d%d%d", &n, &mod, &d, &k) == 4){
        num = 1;
        for(int i = 0; i < n; ++i) scanf("%lld", a+i);
        for(int i = 0; i < n; ++i){
            m[0][i] = i == 0 ? 1 : 0;
            m[1][i] = (i <= d || i >= n-d) ? 1 : 0;
        }
        int x = q_m(k);
        for(int i = 0; i < n; ++i){
            ll sum = 0;
            for(int j = 0; j < n; ++j){
                sum = (sum + (a[j]*m[x][(n-i+j)%n])%mod) % mod;
            }
            if(i != n-1) printf("%lld ", sum);
            else printf("%lld
", sum);
        }
    }
    return 0;
}

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 http://acm.hdu.edu.cn/showproblem.php?pid=2256

盗图

（/2 + /3）^2*n  = = (5+2/6)^n;

(5+2/6)^n + (5-2/6)^n = 1 

#include <iostream>
#include <cmath>
using namespace std;
struct Node{
    int x[2][2];
};
int n;
const int MOD = 1024;
Node mul(Node x, Node y){
    Node X;
    X.x[0][0] = X.x[0][1] = X.x[1][0] = X.x[1][1] = 0;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            for(int k = 0; k < 2; ++k){
                X.x[i][j] = (X.x[i][j] + x.x[i][k]*y.x[k][j]%MOD)%MOD;
            }
        }
    }
    return X;
}
Node q_m(Node A, int x){
    Node AA;
    AA.x[0][0] = AA.x[1][1] = 1;
    AA.x[0][1] = AA.x[1][0] = 0;
    while(x){
        if(x&1) AA = mul(AA, A);
        A = mul(A, A);
        x >>= 1;
    }
    return AA;
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%d", &n);
        Node A;
        A.x[0][0] = A.x[1][1] = 5;
        A.x[0][1] = 12, A.x[1][0] = 2;
        Node AA = q_m(A, n);
        //printf("%d
", (AA.x[0][0]+(int)floor((double)AA.x[1][0]*sqrt(6.0)))%MOD);
        printf("%d
", (2*AA.x[0][0]-1)%MOD);
    }
    return 0;
}

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 http://acm.hdu.edu.cn/showproblem.php?pid=4565

 套路题

（X，Y，Z）|A| = （X', Y', Z'）

#include <iostream>
using namespace std;
#define ll long long
ll a, b, n, m;
struct Node{
    ll x[2][2];
};
Node mul(Node xx, Node yy){
    Node AAA;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            AAA.x[i][j] = 0;
            for(int k = 0; k < 2; ++k){
                AAA.x[i][j] = (AAA.x[i][j] + xx.x[i][k]*yy.x[k][j]%m)%m;
            }
        }
    }
    return AAA;
}
Node q_m(Node A, int x){
    Node AA;
    AA.x[0][0] = AA.x[1][1] = 1;
    AA.x[0][1] = AA.x[1][0] = 0;
    while(x){
        if(x&1) AA = mul(AA, A);
        A = mul(A, A);
        x >>= 1;
    }
    return AA;
}
int main(){
    while(~scanf("%lld%lld%lld%lld", &a, &b, &n, &m)){
        Node A;
        A.x[0][0] = a, A.x[0][1] = 1;
        A.x[1][0] = b, A.x[1][1] = a;
        Node AA = q_m(A, n);
        //cout<<AA.x[0][0]<<endl;
        printf("%lld
", 2*(AA.x[0][0])%m);
    }
    return 0;
}

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http://acm.hdu.edu.cn/showproblem.php?pid=4291

矩阵幂次幂，，找循环节。。。。。。看别人的代码才相通的，还是自己太菜

#include <iostream>
using namespace std;
#define ll long long
ll Mod;
ll Mo[3] = {183120LL, 222222224LL, 1000000007LL};
ll a[3];
ll call(ll x){
    a[0] = 0LL, a[1] = 1LL, a[2] = 3LL;
    for(ll i = 3LL; ;i++){
        a[i%3] = (3LL*a[(i-1)%3]+a[(i-2)%3])%x;
        if(a[i%3] == 3LL && a[(i-1)%3] == 1LL) return i-2;
    }
}
ll n;
struct Node{
    ll x[2][2];
};
Node mul(Node xx, Node yy){
    Node AAA;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            AAA.x[i][j] = 0;
            for(int k = 0; k < 2; ++k){
                AAA.x[i][j] = (AAA.x[i][j] + xx.x[i][k]*yy.x[k][j]%Mod)%Mod;
            }
        }
    }
    return AAA;
}
ll q_m(ll x){
    for(int i = 0; i < 3; ++i){
        Mod = Mo[i];
        Node AA, A;
        AA.x[0][0] = AA.x[1][1] = 1;
        AA.x[0][1] = AA.x[1][0] = 0;
        A.x[0][0] = 3, A.x[0][1] = 1;
        A.x[1][0] = 1, A.x[1][1] = 0;
        while(x){
            if(x&1) AA = mul(AA, A);
            A = mul(A, A);
            x >>= 1;
        }
        x = AA.x[0][0]-1;
        if(x == -1) return 0;
    }
    return x+1;
}
int main(){
    while(~scanf("%lld", &n)){
        if(n == 0){
            puts("0");
            continue;
        }
        //Node AA = q_m(A, n-1);
        //cout<<AA.x[0][0]<<endl;
        printf("%lld
", q_m(n-1));
        //cout<<call(Mod)<<endl;//222222224
        //cout<<call(222222224LL)<<endl;//183120
    }
    return 0;
}

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http://poj.org/problem?id=3233

求矩阵一次方加到k次方对m取模

等差数列用二分

#include <stdio.h>
#include <iostream>
#include <string.h>
using namespace std;
#define ll long long
int n, k, m;
struct Node{
    int x[31][31];
}A;
Node Sum(Node a, Node b){
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            a.x[i][j] = (a.x[i][j] + b.x[i][j])%m;
        }
    }
    return a;
}
Node mul(Node a, Node b){
    Node B;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            B.x[i][j] = 0;
            for(int k = 0; k < n; ++k){
                B.x[i][j] = (B.x[i][j] + a.x[i][k]*b.x[k][j]%m)%m;
            }
        }
    }
    return B;
}
Node q_m(Node AAA, Node A, ll x){
    while(x){
        if(x&1) AAA = mul(AAA, A);
        A = mul(A, A);
        x >>= 1;
    }
    return AAA;
}
void solve(){
    Node AA, AAA, X;
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            AA.x[i][j] = 0;
            X.x[i][j] = AAA.x[i][j] = (i == j);
        }
    }
    while(k > 1){
        if(k&1){
            AA = Sum(AA, mul(X, q_m(AAA, A, k))); //开始q_m没有乘X导致WA了好几次
        }
        k >>= 1;
        X = mul(Sum(AAA, q_m(AAA, A, k)), X);
    }
    X = Sum(mul(A, X), AA);
    for(int i = 0; i < n; ++i){
        for(int j = 0; j < n; ++j){
            printf("%d%c", X.x[i][j], (j == n-1 ? '
' : ' '));
        }
    }
}
int main(){
    while(~scanf("%d%d%d", &n, &k, &m)){
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                scanf("%d", &A.x[i][j]);
            }
        }
        solve();
    }
    return 0;
}

 构造矩阵，，woc，这也能构造？？？？，，，厉害了。。。。

设　　　　|A　A| 为X

　　　　　|0　I |

则

 （I, 0） X = (A, A);　　(A, A) X = (A^2, A^2+A)........

#include <iostream>
#include <string.h>
using namespace std;
const int maxn = 62;
int A[maxn][maxn], B[maxn][maxn];
int n, k, m, x;
void mul(int a[maxn][maxn], int b[maxn][maxn], int x){
    int y[maxn][maxn];
    for(int i = 0; i < x; ++i){
        for(int j = 0; j < x; ++j){
            y[i][j] = 0;
            for(int k = 0; k < x; ++k){
                y[i][j] = (y[i][j] + a[i][k]*b[k][j]%m)%m;
            }
        }
    }
    memcpy(a, y, sizeof(y));
}
void q_m(int x, int xx){
    for(int i = 0; i < x; ++i){
        for(int j = 0; j < x; ++j){
            B[i][j] = (i == j);
        }
    }
    while(xx){
        if(xx&1) mul(B, A, x);
        mul(A, A, x);
        xx >>= 1;
    }
    for(int i = 0; i < n; ++i){
        for(int j = n; j < x; ++j){
            printf("%d%c", B[i][j], (j == x-1 ? '
' : ' '));
        }
    }
}
int main(){
    while(~scanf("%d%d%d", &n, &k, &m)){
        memset(A, 0, sizeof(A));
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                scanf("%d", &x);
                A[i][j] = A[i][n+j] = x;
            }
            A[i+n][i+n] = 1;
        }
        q_m(2*n, k);
    }
    return 0;
}

 下面的代码是二分写的，上面是构造矩阵的。很明显，，构造的代码不但简单好写而且耗时短，比二分快了将近三倍。。。6666666

看来还是构造吊一点。。毕竟一次快速幂就可以出来，而二分的话要多次快速幂。。。。

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http://acm.hdu.edu.cn/showproblem.php?pid=5667

矩阵快速幂加费马定理

#include <iostream>
using namespace std;
#define ll long long
ll n, a, b, c, p, Mod;
struct Node{
    ll x[3][3];
};
Node mul(Node a, Node b){
    Node B;
    for(int i = 0; i < 3; ++i){
        for(int j = 0; j < 3; ++j){
            B.x[i][j] = 0;
            for(int k = 0; k < 3; ++k){
                B.x[i][j] = (B.x[i][j] + a.x[i][k]*b.x[k][j]%Mod)%Mod;
            }
        }
    }
    return B;
}
ll Q_m(ll a, ll x){
    if(a == 0) return 0;
    ll X = 1LL;
    while(x){
        if(x&1) X = X*a%p;
        a = a*a%p;
        x >>= 1;
    }
    return X;
}
void q_m(){
    Node A, AA;
    Mod = p-1LL;
    for(int i = 0; i < 3; ++i){
        for(int j = 0; j < 3; ++j){
            AA.x[i][j] = (i == j);
            A.x[i][j] = 0LL;
        }
    }
    A.x[1][1] = c%Mod;
    A.x[1][0] = A.x[2][1] = A.x[0][1] = A.x[2][2] = 1LL;
    n --;
    while(n){
        if(n&1) AA = mul(AA, A);
        A = mul(A, A);
        n >>= 1;
    }
    /*
    for(int i = 0; i < 3; ++i){
        for(int j = 0; j < 3; ++j){
            printf("%lld%c", AA.x[i][j], (j == 2 ? '
' : ' '));
        }
    }
    */
    b%=Mod;
    printf("%lld
", Q_m(a%p, ((AA.x[1][0]+AA.x[2][0])%Mod)*b%Mod));
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%lld%lld%lld%lld%lld", &n, &a, &b, &c, &p);
        q_m();
    }
    return 0;
}

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http://acm.hdu.edu.cn/showproblem.php?pid=5451

一个网络赛的题目，，，漩斌大神可是十分钟就A的题，。。。我太菜！！！！！！！！！

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
#define ll long long
ll n, m, Mod;
int a[50000];
bool vis[50000];
struct Node{
    ll x[2][2];
};
int w = 1;
ll Q_m(){
    ll X = 1LL, x = 2LL;
    while(n){
        if(n&1) X = X*x%Mod;
        x = x*x%Mod;
        n >>= 1;
    }
    return X;
}
Node mul(Node a, Node b){
    Node B;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            B.x[i][j] = 0;
            for(int k = 0; k < 2; ++k){
                B.x[i][j] = (B.x[i][j] + a.x[i][k]*b.x[k][j]%Mod)%Mod;
            }
        }
    }
    return B;
}
ll get(Node A, Node AA){
    for(ll i = 1; ; ++i){
     //cout<<"*********"<<endl;
        AA = mul(A, AA);
        if(AA.x[0][0] == 5%Mod && AA.x[0][1] == 2%Mod && AA.x[1][0] == 12%Mod && AA.x[1][1] == 5%Mod){
           // cout<<i<<"***************"<<endl;
            return i;
        }
    }
}
void q_m(){
    Node A, AA;
    AA.x[0][0] = AA.x[1][1] = 1LL;
    AA.x[1][0] = AA.x[0][1] = 0;
    A.x[0][0] = A.x[1][1] = 5LL;
    A.x[0][1] = 2LL, A.x[1][0] = 12LL;
    if(!vis[m]){
        Mod = m;
        Mod = get(A, A);
        vis[m] = true;
        a[m] = Mod;
    }else{
        Mod = a[m];
    }
    ll x = Q_m()+1LL;
    Mod = m;
    while(x){
        if(x&1) AA = mul(AA, A);
        A = mul(A, A);
        x >>= 1;
    }
    printf("Case #%d: %lld
", w++, (2LL*AA.x[0][0]%Mod-1LL+Mod)%Mod);
}
int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        scanf("%lld%lld", &n, &m);
        q_m();
    }
    return 0;
}

该题循环节为 （m+1）*（m-1）

细节参考

关于斐波那契数列的循环节：http://blog.csdn.net/acdreamers/article/details/25616461

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中南邀请赛（湘潭） http://202.197.224.59/OnlineJudge2/index.php/Problem/read/id/1243

找循环节。。

挖了好次，有几个特例。。。。

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;
#define ll long long
struct Node{
    int x[2][2];
};
Node T, M;
int Mod;
char str[100009];
Node mul(Node a, Node b){
    Node c;
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            c.x[i][j] = 0;
            for(int k = 0; k < 2; ++k){
                c.x[i][j] = (c.x[i][j] + a.x[i][k]*b.x[k][j]%7)%7;
            }
        }
    }
    return c;
}
void read(Node b){
    for(int i = 0; i < 2; ++i){
        for(int j = 0; j < 2; ++j){
            if(j == 1) printf("%d
", b.x[i][j]);
            else printf("%d ", b.x[i][j]);
        }
    }
}
bool go(){
    //read(T);
    for(int i = 2; ; ++i){
        //cout<<"&&"<<endl;
        M = mul(T, M);
        if(!M.x[0][0] && !M.x[0][1] && !M.x[1][0] &&!M.x[1][1]){
            //read(T);
            //printf("%d******
", i);
            return false;
        }
        if(M.x[0][0] == T.x[0][0] && M.x[0][1] == T.x[0][1] && M.x[1][0] == T.x[1][0] && M.x[1][1] == T.x[1][1]){
            Mod = i - 1;
            //printf("%d%%%%
", i);
            return true;
        }
    }
}
void q(Node a, ll x){
    Node b;
    b.x[0][0] = b.x[1][1] = 1;
    b.x[0][1] = b.x[1][0] = 0;
    while(x){
        if(x&1) b = mul(b, a);
        a = mul(a, a);
        x >>= 1;
    }
    read(b);
}

void lala(int x){
    if(x == 4){
        M.x[0][0] = T.x[0][0], M.x[0][1] = T.x[0][1];
        M.x[1][0] = T.x[1][0], M.x[1][1] = T.x[1][1];
        go();
        return ;
    }
    for(int i = 0; i < 7; ++i){
        //cout<<i<<"**"<<x<<endl;
        T.x[x/2][x%2] = i;
        lala(x+1);
    }
}

int main(){
    //lala(0);
    while(~scanf(" %s", str)){
        int x, y, z, w;
        scanf("%d%d%d%d", &x, &y, &z, &w);
        T.x[0][0] = M.x[0][0] = x, T.x[0][1] = M.x[0][1] = y;
        T.x[1][0] = M.x[1][0] = z, T.x[1][1] = M.x[1][1] = w;
        int len = strlen(str);
        if(!go()){
            if(len == 1 && str[0] == '1') printf("%d %d
%d %d
", x, y, z, w);
            else printf("0 0
0 0
");
            continue;
        }
        //printf("%d****
", Mod);
        if(Mod == 1){
            printf("%d %d
%d %d
", x, y, z, w);
            continue;
        }
        ll sum = 0;
        for(int i = 0; i < len; ++i){
            //cout<<sum<<" "<<i<<endl;
            sum = (sum*10+str[i]-'0')%Mod;
        }
        //cout<<sum<<endl;
        q(T, sum);
    }
    return 0;
}

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只有不断学习才能进步！