常见的构造方法

POJ3233 Matrix power series

#include<cstdio>
#include<cstring>
using namespace std;
int n,k,P,i,j;
struct M{
    int a[31][31];
    friend M operator*(M x,M y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                for (int k=0;k<n;k++) z.a[i][j]=(z.a[i][j]+x.a[i][k]*y.a[k][j])%P;
        return z;
    }
    friend M operator+(M x,M y){
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++) x.a[i][j]=(x.a[i][j]+y.a[i][j])%P;
        return x;
    }
    friend M operator^(M x,int y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++) z.a[i][i]=1;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A,s;
M solve(int x){
    if (x==1) return A;
    M ans=solve(x/2);
    if (x&1){
        M a=A^(x/2+1);
        return ans*a+ans+a;
    }
    return ans*(A^(x/2))+ans;
}
int main(){
    while (~scanf("%d%d%d",&n,&k,&P)){
        for (i=0;i<n;i++)
            for (j=0;j<n;j++) scanf("%d",&A.a[i][j]);
        s=solve(k);
        for (i=0;i<n;i++)
            for (j=0;j<n;j++) printf("%d%c",s.a[i][j],j==n-1?'\n':' ');
    }
}

hdu1588 Gauss Fibonacci

//https://blog.csdn.net/clover_hxy/article/details/51476556
#include<bits/stdc++.h>
using namespace std;
int n,k,P,i,j,b,nn;
struct M{
    int a[2][2];
    friend M operator*(M x,M y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                for (int k=0;k<n;k++) z.a[i][j]=(z.a[i][j]+(long long)x.a[i][k]*y.a[k][j])%P;
        return z;
    }
    friend M operator+(M x,M y){
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++) x.a[i][j]=(x.a[i][j]+y.a[i][j])%P;
        return x;
    }
    friend M operator^(M x,int y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++) z.a[i][i]=1;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A,s,B,f;
M solve(int x){
    if (x==1) return A;
    M ans=solve(x/2);
    if (x&1){
        M a=A^(x/2+1);
        return ans*a+ans+a;
    }
    return ans*(A^(x/2))+ans;
}
int main(){
    B.a[0][1]=B.a[1][0]=B.a[1][1]=1;n=2;
    while (~scanf("%d%d%d%d",&k,&b,&nn,&P)){
        f.a[0][0]=0;f.a[0][1]=1;
        f=f*(B^b);A=B^k;
        s=f+f*solve(nn-1);
        printf("%d\n",s.a[0][0]);
    }
}

HDu4686 Arc of Dream

//https://blog.csdn.net/dyx404514/article/details/10123695
#include<bits/stdc++.h>
using namespace std;
const int P=1e9+7;
typedef long long ll;
ll n,ax,ay,bx,by;
struct M{
    ll a[5][5];
    friend M operator*(M x,M y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<5;i++)
            for (int j=0;j<5;j++)
                for (int k=0;k<5;k++) z.a[i][j]=(z.a[i][j]+x.a[i][k]*y.a[k][j])%P;
        return z;
    }
    friend M operator^(M x,ll y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<5;i++) z.a[i][i]=1;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A,B;
int main(){
    while (~scanf("%lld",&n)){
        memset(A.a,0,sizeof(A.a));
        scanf("%d%d%d%d%d%d",&A.a[0][1],&ax,&ay,&A.a[0][2],&bx,&by);
        A.a[0][3]=A.a[0][1]*A.a[0][2]%P;A.a[0][4]=1;
        memset(B.a,0,sizeof(B.a));
        B.a[0][0]=B.a[3][0]=B.a[4][4]=1;
        B.a[1][1]=ax;B.a[4][1]=ay;
        B.a[2][2]=bx;B.a[4][2]=by;
        B.a[1][3]=ax*by%P;B.a[2][3]=ay*bx%P;
        B.a[3][3]=ax*bx%P;B.a[4][3]=ay*by%P;
        A=A*(B^n);
        printf("%d\n",A.a[0][0]);
    }
}

hdu2371 题目链

#include<bits/stdc++.h>
using namespace std;
char s[81];
int n,k,i,a[81],b[81];
struct M{
    int a[81][81];
    friend M operator*(M x,M y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                for (int k=0;k<n;k++) z.a[i][j]+=x.a[i][k]*y.a[k][j];
        return z;
    }
    friend M operator^(M x,int y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++) z.a[i][i]=1;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A,B;
int main(){
    while (1){
        scanf("%d%d",&n,&k);
        if (!n && !k) return 0;
        for (i=0;i<n;i++) scanf("%d",&a[i]),a[i]--;
        getchar();
        gets(s);
        for (i=0;i<n;i++) b[a[i]]=i;
        memset(A.a,0,sizeof(A.a));
        for (i=0;i<n;i++) A.a[b[i]][i]=1;
        for (i=0;i<n;i++) B.a[0][i]=i;
        B=B*(A^k);
        for (i=0;i<n;i++) putchar(s[B.a[0][i]]);
        putchar('\n');
    }
}

hdu2157 How many ways??

//https://blog.csdn.net/clove_unique/article/details/51482736
#include<bits/stdc++.h>
using namespace std;
const int P=1000;
int n,m,T,x,y,k;
struct M{
    int a[21][21];
    friend M operator*(M x,M y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                for (int k=0;k<n;k++) z.a[i][j]=(z.a[i][j]+x.a[i][k]*y.a[k][j])%P;
        return z;
    }
    friend M operator^(M x,int y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<n;i++) z.a[i][i]=1;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A,B;
int main(){
    while (1){
        scanf("%d%d",&n,&m);
        if (!n && !m) return 0;
        memset(A.a,0,sizeof(A.a));
        while (m--) scanf("%d%d",&x,&y),A.a[x][y]=1;
        scanf("%d",&T);
        while (T--){
            scanf("%d%d%d",&x,&y,&k);
            B=A^k;
            printf("%d\n",B.a[x][y]);
        }
    }
}

poj3613 Cow Relays

//仿照floyd 
#include<cstdio>
#include<cstring>
#include<map>
using namespace std;
map<int,int>mp;
int k,T,s,t,l,x,y,n;
struct M{
    int a[202][202];
    friend M operator*(M x,M y){
        M z;
        memset(z.a,63,sizeof(z.a));
        for (int i=0;i<n;i++)
            for (int j=0;j<n;j++)
                for (int k=0;k<n;k++) z.a[i][j]=min(z.a[i][j],x.a[i][k]+y.a[k][j]);
        return z;
    }
    friend M operator^(M x,int y){
        M z=x;y--;//运算符重载过了，z不能初始化为单位矩阵 
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A;
int main(){
    scanf("%d%d%d%d",&k,&T,&s,&t);
    memset(A.a,63,sizeof(A.a));
    while (T--){
        scanf("%d%d%d",&l,&x,&y);
        if (!mp[x]) mp[x]=++n;
        if (!mp[y]) mp[y]=++n;
        A.a[mp[x]-1][mp[y]-1]=A.a[mp[y]-1][mp[x]-1]=l;
    }
    A=A^k;
    printf("%d",A.a[mp[s]-1][mp[t]-1]);
}

HDU 3306 Another kind of Fibonacci

//https://www.cnblogs.com/KirisameMarisa/p/4187670.html
#include<bits/stdc++.h>
using namespace std;
const int P=10007;
int n,x,y;
struct M{
    int a[4][4];
    friend M operator*(M x,M y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<4;i++)
            for (int j=0;j<4;j++)
                for (int k=0;k<4;k++) z.a[i][j]=(z.a[i][j]+x.a[i][k]*y.a[k][j])%P;
        return z;
    }
    friend M operator^(M x,int y){
        M z;
        memset(z.a,0,sizeof(z.a));
        for (int i=0;i<4;i++) z.a[i][i]=1;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}f,A;
int main(){
    while (~scanf("%d%d%d",&n,&x,&y)){
        x%=P;y%=P;
        memset(A.a,0,sizeof(A.a));
        memset(f.a,0,sizeof(f.a));
        A.a[0][0]=A.a[1][2]=1;
        A.a[1][3]=x;A.a[3][3]=y;
        A.a[1][0]=A.a[1][1]=x*x%P;
        A.a[2][0]=A.a[2][1]=y*y%P;
        A.a[3][0]=A.a[3][1]=2*x*y%P;
        f.a[0][0]=2;f.a[0][1]=f.a[0][2]=f.a[0][3]=1;
        f=f*(A^(n-1));
        printf("%d\n",f.a[0][0]);
    }
}

HDU5171

/*program1利用二分计算矩阵的和 */
#include<bits/stdc++.h>
using namespace std;
const int P=10000007;
typedef long long ll;
const int N=100002;
int n,k,i,a[N];
ll sum;
struct M{
    ll a[2][2];
    friend M operator*(M x,M y){
        M z;
        for (int i=0;i<2;i++)
            for (int j=0;j<2;j++) z.a[i][j]=(x.a[i][0]*y.a[0][j]+x.a[i][1]*y.a[1][j])%P;
        return z;
    }
    friend M operator+(M x,M y){
        for (int i=0;i<2;i++)
            for (int j=0;j<2;j++) x.a[i][j]=(x.a[i][j]+y.a[i][j])%P;
        return x;
    }
    friend M operator^(M x,int y){
        M z;
        z.a[0][0]=z.a[1][1]=1;
        z.a[0][1]=z.a[1][0]=0;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A,s,f;
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline int read(){
    int x=0,fl=1;char ch=gc();
    for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
    for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
    return x*fl;
}
M solve(int x){
    if (x==1) return A;
    M ans=solve(x/2);
    if (x&1){
        M a=A^(x/2+1);
        return ans*a+ans+a;
    }
    return ans*(A^(x/2))+ans;
}
int main(){
    while (~scanf("%d%d",&n,&k)){
        sum=0;
        for (i=0;i<n;i++) a[i]=read(),sum+=a[i];
        nth_element(a,a+n-2,a+n);
        f.a[0][0]=a[n-2];
        f.a[0][1]=a[n-1];   
        A.a[0][1]=A.a[1][0]=A.a[1][1]=1;
        f.a[1][0]=f.a[1][1]=A.a[0][0]=0;
        s=f*solve(k);
        printf("%lld\n",(sum+s.a[0][1])%P);
    }
}

/*program2 思路来自：https://blog.csdn.net/crescent__moon/article/details/43617703但不尽相同 当f[n]=f[n-1]+f[n-2]时，s[n]=f[n+2]-f[1]可用数学归纳法证明 */
#include<bits/stdc++.h>
using namespace std;
const int P=10000007;
typedef long long ll;
const int N=100002;
int n,k,i,a[N];
ll sum;
struct M{
    ll a[2][2];
    friend M operator*(M x,M y){
        M z;
        for (int i=0;i<2;i++)
            for (int j=0;j<2;j++) z.a[i][j]=(x.a[i][0]*y.a[0][j]+x.a[i][1]*y.a[1][j])%P;
        return z;
    }
    friend M operator^(M x,int y){
        M z;
        z.a[0][0]=z.a[1][1]=1;
        z.a[0][1]=z.a[1][0]=0;
        for (;y;y>>=1,x=x*x)
            if (y&1) z=z*x;
        return z;
    }
}A,f;
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline int read(){
    int x=0,fl=1;char ch=gc();
    for (;ch<48||ch>57;ch=gc())if(ch=='-')fl=-1;
    for (;48<=ch&&ch<=57;ch=gc())x=(x<<3)+(x<<1)+(ch^48);
    return x*fl;
}
int main(){
    while (~scanf("%d%d",&n,&k)){
        sum=0;
        for (i=0;i<n;i++) a[i]=read(),sum+=a[i];//其实应该加到n-3的，多把a[n-2]和a[n-1]加进去了，要减掉 
        nth_element(a,a+n-2,a+n);
        f.a[0][0]=a[n-2];
        f.a[0][1]=a[n-1];
        A.a[0][1]=A.a[1][0]=A.a[1][1]=1;
        f.a[1][0]=f.a[1][1]=A.a[0][0]=0;
        f=f*(A^(k+2));
        printf("%lld\n",(sum+f.a[0][1]-a[n-1]-a[n-1]-a[n-2])%P);
    }
}