nefu暑假acm集训1 构造矩阵个人模板+例题汇总

前言：

以下都是nefu暑假集训的训练题，我在此把我的模板和写的一些练习题汇总一下并分享出来，希望在能满足我复习的情况下能帮助到你。

正文：

模板：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=mod;
typedef struct{ll m[n][n];//一个n*n的矩阵，注意矩阵不能太大，若太大则不能给他写成结构 
}; 
matrix P={//内部填写自己经过推导构造出来的矩阵 
};
matrix I={//内部填写自相应大小的单位矩阵 
};
matrix mul(matrix a,matrix b){//对两个大小为n*n的矩阵进行相乘并取模操作 matrix c;for(int i=0;i<n;i++){for(int j=0;j<n;j++){c.m[i][j]=0;for(int k=0;k<n;k++){a.m[i][k]=(a.m[i][k]%mod+mod)%mod; b.m[k][j]=(b.m[k][j]%mod+mod)%mod; c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;}c.m[i][j] = (c.m[i][j]%mod+mod)%mod;}}return c;
}
matrix quickpow(ll n){//对矩阵进行快速幂操作 matrix m=P,b=I;while(n){if(n&1)b=mul(b,m);n>>=1;m=mul(m,m);}return b;
}

这是矩阵连乘的基础模板

习题：

1、Not Fibonacci：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e7;
typedef struct{ll m[3][3];
}matrix;
matrix P={0,0,0,1,0,0,0,0,1
};
matrix I={1,0,0,0,1,0,0,0,1
};
matrix mul(matrix a,matrix b){matrix c;for(int i=0;i<3;i++){for(int j=0;j<3;j++){c.m[i][j]=0;for(int k=0;k<3;k++){a.m[i][k]=(a.m[i][k]%mod+mod)%mod; b.m[k][j]=(b.m[k][j]%mod+mod)%mod; c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;}c.m[i][j] = (c.m[i][j]%mod+mod)%mod;}}return c;
}
matrix quickpow(ll n){matrix m=P,b=I;while(n){if(n&1)b=mul(b,m);n>>=1;m=mul(m,m);}return b;
}
ll a,b,p,q,s,e;
int main(){int t;cin>>t;while(t--){ll l,r;matrix tmp;cin>>a>>b>>p>>q>>s>>e;P.m[0][0]=p;P.m[0][1]=q; P.m[2][0]=p;P.m[2][1]=q;if(e==0)r=(a%mod+mod)%mod;else if(e==1)r=((a+b)%mod+mod)%mod;else{tmp=quickpow(e-1);r=(tmp.m[2][0]*b)%mod+(tmp.m[2][1]*a)%mod+(tmp.m[2][2]*(a+b))%mod;r=(r%mod+mod)%mod;}if(s==0)l=0;else if(s==1)l=a;else if(s==2)l=((a+b)%mod+mod)%mod;else{tmp=quickpow(s-2);l=(tmp.m[2][0]*b)%mod+(tmp.m[2][1]*a)%mod+(tmp.m[2][2]*(a+b))%mod;l=(l%mod+mod)%mod;}cout<<((r-l)%mod+mod)%mod<<endl;}return 0;
}

构造出相应的p矩阵为 $\begin{vmatrix} p & q& 0\\ 1 &0 &0 \\ p &q &1 \end{vmatrix}$ ，初始矩阵为 $\begin{vmatrix} a_{n}\\ a_{n-1}\\ s_{n}\end{vmatrix}$ ，在根据s,e分别计算即可。

2、Another kind of Fibonacci

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=10007;
typedef struct{ll m[4][4];
}matrix;
matrix P={1,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0
};
matrix I={1,0,0,0,0,1,0,0,0,0,1,0,0,0,0,1
};
matrix mul(matrix a,matrix b){matrix c;for(int i=0;i<4;i++){for(int j=0;j<4;j++){c.m[i][j]=0;for(int k=0;k<4;k++){a.m[i][k]=(a.m[i][k]%mod+mod)%mod; b.m[k][j]=(b.m[k][j]%mod+mod)%mod; c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;}c.m[i][j] = (c.m[i][j]%mod+mod)%mod;}}return c;
}
matrix quickpow(ll n){matrix m=P,b=I;while(n){if(n&1)b=mul(b,m);n>>=1;m=mul(m,m);}return b;
}
ll n,x,y;
int main(){while(cin>>n>>x>>y){ll sum;P.m[0][1]=x*x%mod;P.m[0][2]=2*x*y%mod;P.m[0][3]=y*y%mod;P.m[1][1]=x*x%mod;P.m[1][2]=2*x*y%mod;P.m[1][3]=y*y%mod;P.m[2][1]=x%mod;P.m[2][2]=y%mod;if(n==0)sum=1;else if(n==1)sum=2;else{matrix tmp=quickpow(n-1);sum=(tmp.m[0][0]*2%mod)+(tmp.m[0][1]%mod)+(tmp.m[0][2]%mod)+(tmp.m[0][3]%mod);sum=(sum%mod+mod)%mod;}cout<<sum<<endl;}return 0;
}

上题的稍难版，不过核心还是推矩阵p和初始矩阵，得出来便可以直接代入模板。

3、Fibonacci Number Problem

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn = 54;
LL mod, size;//size = 矩阵大小
typedef struct
{LL m[maxn][maxn];
} Matrix;Matrix P;
Matrix I;
//正常快速幂
LL quick_mod(LL a, LL b, LL c)
{LL ans = 1;if(b == 0)return 1;while(b){if(b & 1)ans = (ans*a)%c;b >>= 1;a = (a*a)%c;}return ans;
}
//矩阵乘法
Matrix matrix_mul(Matrix a, Matrix b)
{int i, j, k;Matrix c;for(i=0; i<size; i++){for(j=0; j<size; j++){c.m[i][j] = 0;for(k=0; k<size; k++){c.m[i][j] += ((a.m[i][k]%mod)*(b.m[k][j]%mod))%mod;}c.m[i][j] %= mod;}}return c;
}
//矩阵的快速幂
Matrix quick_pow(LL m)
{Matrix b=P, ans=I;while(m){if(m & 1)ans = matrix_mul(ans, b);m >>= 1;b = matrix_mul(b, b);}return ans;
}
//组合数。。。
LL c[50][50];
int main()
{memset(c, 0, sizeof(c));for(int i=0; i<50; i++){c[i][0] = 1;c[i][i] = 1;}for(int i=1; i<50; i++)for(int j=1; j<i; j++)c[i][j] = c[i-1][j] + c[i-1][j-1];int t;Matrix tmp;LL f1, f2, a, b, k, n, m ;LL sum, ans1, ans2;//scanf("%d",&t);cin>>t;while(t--){sum = 0;cin>>f1>>f2>>a>>b>>k>>n>>mod;//scanf("%lld%lld%lld%lld%lld%lld%lld",&f1,&f2,&a,&b,&k,&n,&mod);memset(P.m, 0, sizeof(P.m));memset(I.m, 0, sizeof(I.m));if(k == 0)printf("%lld\n",n%mod);else{if(n == 1)cout<<quick_mod(f1, k, mod)<<endl;else if(n == 2)cout<<(quick_mod(f1,k,mod) + quick_mod(f2,k,mod))%mod<<endl;else{size = k+2;//矩阵赋值for(int i=0; i<size; i++)I.m[i][i] = 1;P.m[0][0] = 1;P.m[0][size-1] = 1;for(int i=1; i<size-1; i++)P.m[0][i] = 0;for(int i=1; i<size; i++)for(int j=size-i,w=0; j<size; j++,w++){P.m[i][j]=(((c[i-1][w]%mod)*quick_mod(a,w,mod))%mod)*quick_mod(b,i-1-w,mod);P.m[i][j] %= mod;}tmp = quick_pow(n-1);sum = (sum+(tmp.m[0][0]%mod)*(quick_mod(f1,k,mod)))%mod;for(int i=1; i<size; i++){ans1 = (quick_mod(f1,size-1-i,mod)*quick_mod(f2,i-1,mod))%mod;ans2 = (ans1*(tmp.m[0][i]%mod))%mod;sum = (sum+ans2)%mod;}cout<<sum%mod<<endl;}}}return 0;
}

首先一个难题就是怎么判断矩阵的维数（矩阵的维数是个变量），解决方法：开一个比较大的数组，然后再用一个公有变量记一下就行了，具体详见代码；
矩阵k次方是啥，找规律来求解；

4、A Simple Math Problem

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef struct{ll m[10][10];
}matrix;
matrix P={0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0
};
matrix I={1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1
};
int k,m;
matrix mul(matrix a,matrix b){matrix c;for(int i=0;i<10;i++){for(int j=0;j<10;j++){c.m[i][j]=0;for(int k=0;k<10;k++){a.m[i][k]=(a.m[i][k]%m+m)%m; b.m[k][j]=(b.m[k][j]%m+m)%m; c.m[i][j]+=(a.m[i][k]*b.m[k][j])%m;}c.m[i][j] = (c.m[i][j]%m+m)%m;}}return c;
}
matrix quickpow(ll n){matrix m=P,b=I;while(n){if(n&1)b=mul(b,m);n>>=1;m=mul(m,m);}return b;
}int main(){while(cin>>k>>m){cin>>P.m[0][0]>>P.m[0][1]>>P.m[0][2]>>P.m[0][3]>>P.m[0][4]>>P.m[0][5]>>P.m[0][6]>>P.m[0][7]>>P.m[0][8]>>P.m[0][9];ll sum;if(k<10){sum=k%m;}else{matrix tmp=quickpow(k-9);sum=(tmp.m[0][0]*9)%m+(tmp.m[0][1]*8)%m+(tmp.m[0][2]*7)%m+(tmp.m[0][3]*6)%m+(tmp.m[0][4]*5)%m+(tmp.m[0][5]*4)%m+(tmp.m[0][6]*3)%m+(tmp.m[0][7]*2)%m+(tmp.m[0][8])%m;sum=(sum%m+m)%m;}cout<<sum<<endl;}return 0;
}

矩阵稍大稍微有点麻烦，不过还是可以直接写出来的。

5、Fibs之和

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=1e9;
typedef struct{ll m[3][3];
}matrix;
matrix P={0,0,0,1,0,0,0,0,1
};
matrix I={1,0,0,0,1,0,0,0,1
};
matrix mul(matrix a,matrix b){matrix c;for(int i=0;i<3;i++){for(int j=0;j<3;j++){c.m[i][j]=0;for(int k=0;k<3;k++){a.m[i][k]=(a.m[i][k]%mod+mod)%mod; b.m[k][j]=(b.m[k][j]%mod+mod)%mod; c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;}c.m[i][j] = (c.m[i][j]%mod+mod)%mod;}}return c;
}
matrix quickpow(ll n){matrix m=P,b=I;while(n){if(n&1)b=mul(b,m);n>>=1;m=mul(m,m);}return b;
}
ll a,b,p,q,s,e;
int main(){while(cin>>s>>e){if(s==0&&e==0)break;ll l,r;matrix tmp;P.m[0][0]=1;P.m[0][1]=1; P.m[2][0]=1;P.m[2][1]=1;if(e==0)r=1;else if(e==1)r=2;else{tmp=quickpow(e-1);r=(tmp.m[2][0])%mod+(tmp.m[2][1])%mod+(tmp.m[2][2])%mod;r=(r%mod+mod)%mod;}if(s==0)l=0;else if(s==1)l=1;else if(s==2)l=2;else{tmp=quickpow(s-2);l=(tmp.m[2][0])%mod+(tmp.m[2][1])%mod+(tmp.m[2][2])%mod;l=(l%mod+mod)%mod;}cout<<((r-l)%mod+mod)%mod<<endl;}return 0;
}

和1一摸一样

6、fibs的组合

#include<bits/stdc++.h>
using namespace std;
int main(){int n;while(cin>>n){if(n%3==0){cout<<"yes"<<endl;}	else{cout<<"no"<<endl;}}return 0;
}

找规律

7、fibs的位数

#include<bits/stdc++.h>
using namespace std;
int main()
{long long n,a,b,u,v;while(scanf("%lld%lld%lld%lld%lld",&n,&a,&b,&u,&v)!=EOF){double c=sqrt(u*u+4*v);double x=(u+c)/2.0;double y=(u-c)/2.0;double len=n*log10(x)+log10(b-a*y)-log10(c);printf("%lld\n",(long long)len+1);}return 0;
}

这个是纯概念（而且挺偏的），我感觉比赛不会有这种内容，想了解可以看看这个NEFU 461 fibs的位数()_现在让你计算广义fibonacci数列的位数。-CSDN博客

8、Matrix multiplication

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int MAXN = 817;
const int mod = 3;
int A[MAXN][MAXN], B[MAXN][MAXN];
int C[MAXN][MAXN];
int n;int Scan()
{int res = 0, ch;ch=getchar();if(ch >= '0' && ch <= '9')		 res = ch - '0';while((ch = getchar()) >= '0' && ch <= '9' )res = res * 10 + ch - '0';return res;
}void input()
{int i, j;for(i = 1; i <= n; i++){for(j = 1; j <= n; j++){//scanf("%d",&A[i][j]);//A[i][j] %= mod;A[i][j]=Scan()%3;}}for(i = 1; i <= n; i++){for(j = 1; j <= n; j++){//scanf("%d",&B[i][j]);//B[i][j] %= mod;B[i][j]=Scan()%3;}}
}void multi()
{//两个相等矩阵的乘法，对于稀疏矩阵，有在0处不用运算的优化memset(C,0,sizeof(C));for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(A[i][j] == 0)//稀疏矩阵优化continue;for(int k = 1; k <= n; k++){C[i][k] += A[i][j]*B[j][k];//i行k列第j项//	C[i][k] %= mod;}}}
}void print()//输出矩阵信息
{for(int i = 1; i <= n; i++){for(int j = 1; j <= n; j++){if(j == 1)printf("%d",C[i][j]%mod);elseprintf(" %d",C[i][j]%mod);}printf("\n");}
}int main()
{while(~scanf("%d",&n)){input();multi();print();}return 0;
}

这题题目不难，但卡时间有一点刻意了，十分恶心。

9、Number Sequence

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int mod=7;
typedef struct{ll m[2][2];
}matrix;
matrix P={1,1,1,0
};
matrix I={1,0,0,1
};
matrix mul(matrix a,matrix b){matrix c;for(int i=0;i<2;i++){for(int j=0;j<2;j++){c.m[i][j]=0;for(int k=0;k<2;k++){a.m[i][k]=(a.m[i][k]%mod+mod)%mod; b.m[k][j]=(b.m[k][j]%mod+mod)%mod; c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;}c.m[i][j] = (c.m[i][j]%mod+mod)%mod;}}return c;
}
matrix quickpow(ll n){matrix m=P,b=I;while(n){if(n&1)b=mul(b,m);n>>=1;m=mul(m,m);}return b;
}
int main(){ll a,b,n;while(cin>>a>>b>>n){if(a==0&&b==0&&n==0)break;P.m[0][0]=a;P.m[0][1]=b;ll sum=0;if(n==-1)break;if(n==0){cout<<0<<endl;continue;}if(n==1){cout<<1<<endl;continue;}matrix tmp=quickpow(n-2);sum=(tmp.m[0][0])%mod+(tmp.m[0][1])%mod;sum=(sum%mod+mod)%mod;cout<<sum<<endl;}return 0;
}

模板

10、Fibonacci

#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;
const int mod=10000;
typedef struct{ll m[2][2];
}matrix;
matrix P={1,1,1,0
};
matrix I={1,0,0,1
};
matrix mul(matrix a,matrix b){matrix c;for(int i=0;i<2;i++){for(int j=0;j<2;j++){c.m[i][j]=0;for(int k=0;k<2;k++){a.m[i][k]=(a.m[i][k]%mod+mod)%mod; b.m[k][j]=(b.m[k][j]%mod+mod)%mod; c.m[i][j]+=(a.m[i][k]*b.m[k][j])%mod;}c.m[i][j] = (c.m[i][j]%mod+mod)%mod;}}return c;
}
matrix quickpow(ll n){matrix m=P,b=I;while(n){if(n&1)b=mul(b,m);n>>=1;m=mul(m,m);}return b;
}
int main(){ll n;while(cin>>n){ll sum=0;if(n==-1)break;if(n==0){cout<<0<<endl;continue;}if(n==1){cout<<1<<endl;continue;}matrix tmp=quickpow(n-2);sum=(tmp.m[0][0])%mod+(tmp.m[0][1])%mod;sum=(sum%mod+mod)%mod;cout<<sum<<endl;}return 0;
}

求最后四位就是%10000

11、Gauss Fibonacci

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct matrie{ll m[10][10];
};
int k,b,n,mod;
matrie A,B,T,C,D,E;
matrie multi(matrie a,matrie b){matrie c;memset(c.m,0,sizeof(c.m));for(int i=0;i<10;i++){for(int j=0;j<10;j++){for(int k=0;k<10;k++){c.m[i][j]=(a.m[i][k]*b.m[k][j]+c.m[i][j])%mod;}}}return c;
}
matrie fast_power(matrie a, ll b) {matrie res;memset(res.m,0,sizeof(res.m));for(int i = 0; i < 10; ++i)res.m[i][i] = 1;while(b>0){if(b&1){res=multi(res,a);}b>>=1;a=multi(a, a);}return res;
}
void init(){T.m[0][0]=0;T.m[0][1]=1;T.m[1][0]=1;T.m[1][1]=1;A.m[0][0]=0;A.m[0][1]=1;C=multi(A,fast_power(T,b));E=fast_power(T,k);for(int i=0;i<2;i++){D.m[i][i]=1;B.m[i][i]=B.m[i][i+2]=1;}for(int i=2;i<4;i++){for(int j=0;j<2;j++){D.m[i][j]=D.m[i][j+2]=E.m[i-2][j];}}
}
int main(){while(scanf("%d%d%d%d",&k,&b,&n,&mod)!=EOF){init();matrie tmp=multi(B,fast_power(D,n-1));matrie sum;memset(sum.m, 0, sizeof(sum.m));for(int i=0;i<2;i++){for(int j=0;j<2;j++){sum.m[i][j]=tmp.m[i][j];}}matrie ans=multi(C,sum);printf("%lld\n", ans.m[0][0]);}return 0;
}

详解可以看看这篇博客HDU 1588 Gauss Fibonacci(分块矩阵优化)-CSDN博客

12、Fast Matrix Calculation

#include<bits/stdc++.h>
#define LL long long int 
const int MOD=6;
int n,m;
struct node{LL m[10][10];node(){memset(m,0,sizeof(m));}
};
int a[1005][10],b[10][1005],c[1005][10],d[1005][1005];
node cla(node A,node B)
{node C;for(int i=0;i<m;i++)//A对应的行 for(int j=0;j<m;j++)//B对应的列 for(int k=0;k<m;k++) if(A.m[i][k]&&B.m[k][j])//剪枝（添条件，设门槛），提高效率，有一个是0，相乘肯定是0{C.m[i][j]+=A.m[i][k]*B.m[k][j];C.m[i][j]%=MOD;}return C;
}
node POW(int k,node ans)
{   node e;for(int i=0;i<m;i++) e.m[i][i]=1;while(k){if(k%2) e=cla(e,ans);ans=cla(ans,ans);k/=2;}return e; 
} 
int main()
{while(~scanf("%d%d",&n,&m)){if(n==0&&m==0)break;memset(a,0,sizeof(a));memset(b,0,sizeof(b));memset(c,0,sizeof(c));memset(d,0,sizeof(d));for(int i=0;i<n;i++)for(int j=0;j<m;j++)scanf("%d",&a[i][j]);for(int i=0;i<m;i++)for(int j=0;j<n;j++)scanf("%d",&b[i][j]);node ans;for(int i=0;i<m;i++)//A对应的行 for(int j=0;j<m;j++)//B对应的列 for(int k=0;k<n;k++) if(b[i][k]&&a[k][j])//剪枝（添条件，设门槛），提高效率，有一个是0，相乘肯定是0{ans.m[i][j]+=b[i][k]*a[k][j];ans.m[i][j]%=MOD;}ans=POW(n*n-1,ans);for(int i=0;i<n;i++)//A对应的行 for(int j=0;j<m;j++)//B对应的列 for(int k=0;k<m;k++) if(a[i][k]&&ans.m[k][j])//剪枝（添条件，设门槛），提高效率，有一个是0，相乘肯定是0{c[i][j]+=a[i][k]*ans.m[k][j];c[i][j]%=MOD;}for(int i=0;i<n;i++)//A对应的行 for(int j=0;j<n;j++)//B对应的列 for(int k=0;k<m;k++) if(c[i][k]&&b[k][j])//剪枝（添条件，设门槛），提高效率，有一个是0，相乘肯定是0{d[i][j]+=c[i][k]*b[k][j];d[i][j]%=MOD;}int sum=0;for(int i=0;i<n;i++)//A对应的行 for(int j=0;j<n;j++)//B对应的列 sum+=d[i][j];printf("%d\n",sum); }
}

因为n实在太大了，开不了1000*1000的矩阵，但是可以利用矩阵相乘的结合律将（A*B）^n转换为A*(B*A)^(n-1)*B，这样B和A相乘得到的是一个6*6(最多)矩阵, 将这个矩阵用快速幂计算(n*n-1)次幂，在乘上一开始的A和最后的B即可。