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Re:矩阵快速幂In Reply To:矩阵快速幂 Posted by:ACAccepted at 2019-07-22 09:59:55 > #include <cstdio> > #include <串> 使用命名空间 std >; > >模板<类型名 T> > 内联 T 读取() > { > T x=0,f=1;char c=getchar(); > while (c<'0' || c>'9'){if (c=='-')f=-1;c=getchar();} > while (c>='0'&&c<='9'){x=(x<<1)+(x<<3)+c-48;c=getchar();} >返回 x*f; > } > > const int MAXN=35; > int n,k,mod; > >结构矩阵1 > { > int n,m; > 长 a[MAXN][MAXN]; > > inline matrix1(){n=m=0;memset(a,0,sizeof(a));} > > inline void init(int n,int w) > { > this->n=n;this->m=n; > for (int i=1;i<=n;i++) a[i][i]=w; > } > > 内联无效打印() > { > for (int i=1;i<=n;i++) > { > for (int j=1;j<=m;j++) printf(“%d ”,a[i][j]); > printf(“\n”); > } > } > > 内联结构矩阵 1 运算符 + (结构矩阵 1 TMP) > { > struct matrix1 ans;ans.n=tmp.n;ans.m=tmp.m; > for (int i=1;i<=ans.n;i++) > for (int j=1;j<=ans.m;j++) > ans.a[i][j]=(a[i][j]+tmp.a[i][j])%mod; >返回; > } > >内联结构矩阵 1 运算符 * (结构矩阵 1 TMP) > { >结构矩阵1 ans; > ans.n=n;ans.m=tmp.m; > for (int i=1;i<=ans.n;i++) > for (int j=1;j<=ans.m;j++) > for (int k=1;k<=tmp.n;k++) > ans.a[i][j]=(ans.a[i][j]+a[i][k]*tmp.a[k][j]%mod)%mod; >返回; > } > }A; > >结构矩阵2 > { > int n,m; >结构矩阵1 a[3][3]; > > inline matrix2(){n=m=0;} > > 内联结构矩阵 2 运算符 * (结构矩阵 2 TMP) > { >结构矩阵2 ans; > ans.n=n;ans.m=tmp.m; > for (int i=1;i<=ans.n;i++) > for (int j=1;j<=ans.m;j++) > for (int k=1;k<=tmp.n;k++) > ans.a[i][j]=ans.a[i][j]+a[i][k]*tmp.a[k][j]; >返回; > } > }tmp,f; > > 内联结构矩阵 2 qpow(struct matrix2 a,int b) > { >结构矩阵2 ans; > ans.m=ans.n=2; > ans.a[1][1].init(n,1);ans.a[1][2].init(n,0); > ans.a[2][1].init(n,0);ans.a[2][2].init(n,1); > 同时 (b) > { > if (b&1) ans=ans*a; > a=a*a;b>>=1; > } >返回; > } > > int main() > { > n=read<int>();k=read<int>();mod=read<int>(); > A.n=n;上午=n; > for (int i=1;i<=n;i++) > for (int j=1;j<=n;j++) > A.a[i][j]=read<int>(); > tmp.n=2;tmp.m=2; > tmp.a[1][1]=A;tmp.a[1][2].init(n,1); > tmp.a[2][1].init(n,0);tmp.a[2][2].init(n,1); > f.n=2;f.m=1; > f.a[1][1]=A;f.a[2][1]=A; > tmp=qpow(tmp,k-1); > f=tmp*f; > f.a[1][1].print(); >返回 0; > } Followed by: Post your reply here: |
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