2017 ECJTU ACM程序设计竞赛 矩阵快速幂+二分

矩阵

Time Limit : 3000/1000ms (Java/Other)   Memory Limit : 65535/32768K (Java/Other)

Total Submission(s) : 7   Accepted Submission(s) : 4

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Problem Description

假设你有一个矩阵，有这样的运算A^(n+1) = A^(n)*A (*代表矩阵乘法)
现在已知一个n*n矩阵A,S = A+A^2+A^3+...+A^k,输出S，因为每一个元素太大了，输出的每个元素模10

Input

先输入一个T(T<=10)，每组一个n,k(1<=n<=30, k<=1000000)

Output

输出一个矩阵,每个元素模10(行末尾没有多余空格)

Sample Input

1
3 2
0 2 0
0 0 2
0 0 0

Sample Output

0 2 4
0 0 2
0 0 0

矩阵快速幂 + 等比数列二分求和
AC

 #include <cstdio>
 #include <cmath>
 #include <cstring>
 #include <cstdlib>
 #include <iostream>
 #include <sstream>
 #include <algorithm>
 #include <string>
 #include <queue>
 #include <vector>
 using namespace std;
 const int maxn= 1e5+;
 const double eps= 1e-;
 const int inf = 0x3f3f3f3f;
 const int mod =;
 typedef long long ll;
 int n,m;
 struct matrix
 {
     int m[][];
     matrix()
     {
         memset(m,,sizeof(m));
     }
 };
 matrix operator *(const matrix &a,const matrix &b)
 {
     matrix c;
     for(int i=;i<=n;i++)
         for(int j=;j<=n;j++)
         {
             c.m[i][j]=;
             for(int k=;k<=n;k++)
                 c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
         }
         return c;
 }
 matrix quick(matrix base,int pow)
 {
     matrix a;
     for(int i=;i<=n;i++) a.m[i][i]=;
     while(pow)
     {
         if(pow&) a=a*base;
         base=base*base;
         pow>>=;
     }
     return a;
 }
 matrix sum(matrix a,matrix b)
 {
     for(int j=;j<=n;j++)
         for(int k=;k<=n;k++)
             a.m[j][k]=(a.m[j][k]+b.m[j][k])%mod;
     return a;
 }
 matrix solve(matrix x,int y)
 {
     matrix a,s;
     if(y==)
         return x;
     a=solve(x,y/);
     s=sum(a,a*quick(x,y/));
     if(y&)
         s=sum(s,quick(x,y));
     return s;
 }
 int main()
 {
     int t;
     scanf("%d",&t);
     while(t--)
     {
         scanf("%d %d",&n,&m);
         matrix  a;
         for(int i=;i<=n;i++)
             for(int j=;j<=n;j++)
                 scanf("%d",&a.m[i][j]);
         matrix ans=solve(a,m);
         for(int i=;i<=n;i++)
         {
             for(int j=;j<=n;j++)
             {
                 if(j==n)
                     printf("%d\n",ans.m[i][j]);
                 else
                     printf("%d ",ans.m[i][j]);
             }
         }

     }
 }