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Custom Home Remodeling & Roofing has been in business for almost 10 years now and is founded on the pillars of trust, honesty and professionalism in an industry that is fraught with a lack of it!
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As the seasons change and time passes by, the exterior of your home will be affected by the elements. Especially your roof. Are you already noticing that something might be wrong with it? It’s your lucky day, because CHR Roofing is here to make everything right. We are Southern CA based roofing contractor that knows what your residence needs. From shingle work to complete roof installations, we do it all! For more than 10 years, our licensed, bonded, and insured company has helped people in the area. If you are looking for high quality CHR Roofing services, you have come to the right place!
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What sets CHR Roofing apart from your regular roofing company is not only the variety of our roofing services, but our impeccable quality control protocols. After all, you wish to have a roof that will withstand rain, wind, and hale. In order for your roof to be as durable and long lasting as possible, our specialists always make an extra effort and use only the finest products and materials available on the market. Give us the chance to show you what our roofing professionals can do, and you will be more than impressed!
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Thumbtack Award> poj 3233 Matrix Power Series 矩阵高速幂
poj 3233 Matrix Power Series 矩阵高速幂
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poj 3233 Matrix Power Series 矩阵快速幂
　　/*愧疚了，好久没写什么有意思的题了。
这道基于二分的矩阵连乘，方法很经典。
求幂二分，求和再二分。经典。
求幂的时候，若奇数，A^n=A^(n/2)*A^(n/2)*A.
偶数时 A^n=A^(n/2)*A^(n/2)。
求和的时候，若为奇数，A^1+A^2+...+A^(2*m+1)=(A^1+...+A^m)+(A^1+...+A^m)*A^m*A,
A^1+A^2+...+A^(2*m+1)=(A^1+...+A^m)+(A^1+...+A^m)*A^m。
奇数的时候，构造0,1单位矩阵。
对奇数进行特判。
memcpy(a,b,sizeof(int)*N*N);
切记memcpy的赋值范围。之前因为太小，答案一直错误。*/
#include &stdio.h&
#include &cstring&
int n,m,k;
const int N=31;
int ans[N][N],A[N][N];
void add(int a[][N],int b[][N])
for(int i=0; i&n; i++)
for(int j=0; j&n; j++)
a[i][j]+=b[i][j];
if(a[i][j]&=m) a[i][j]%=m;
void mul(int a[][N],int b[][N])
int tmp[N][N];
for(int i=0; i&n; i++)
for(int j=0; j&n; j++)
tmp[i][j]=0;
for(int k=0; k&n; k++)
tmp[i][j]+=a[i][k]*b[k][j];
if(tmp[i][j]&=m) tmp[i][j]%=m;
memcpy(a,tmp,sizeof(tmp));
void pow(int a[][N],int b[][N],int x)
int tmp[N][N];
memcpy(tmp,b,sizeof(int)*N*N);
for(int i=0; i&n; i++)
for(int j=0; j&n; j++)
if(i==j) a[i][j]=1;
else a[i][j]=0;
if(x&1)//最后的结果存在数组a中，因为最后x比为1，符合条件，自己模拟一遍则易知.
mul(a,tmp);
mul(tmp,tmp);
void mat_sum(int a[][N],int b[][N],int x)
int c[N][N],d[N][N];
memset(a,0,sizeof(int)*N*N);
memcpy(a,b,sizeof(int)*N*N);
mat_sum(a,b,x$>$1);
pow(c,b,x$>$1);
memcpy(d,a,sizeof(int)*N*N);
int main()
while(scanf(&%d%d%d&,&n,&k,&m)==3)
for(int i=0; i&n; i++)
for(int j=0; j&n; j++)
scanf(&%d&,&A[i][j]);
mat_sum(ans,A,k);
for(int i=0; i&n; i++)
printf(&%d&,ans[i][0]);
for(int j=1; j&n; j++)
if(j&n-1) printf(& %d&,ans[i][j]);
else printf(& %d\n&,ans[i][j]);
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