2017 ECJTU ACM程序设计竞赛 矩阵快速幂+二分

Posted 2020-10-15 灬从此以后灬

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

 1 #include <cstdio>
 2 #include <cmath>
 3 #include <cstring>
 4 #include <cstdlib>
 5 #include <iostream>
 6 #include <sstream>
 7 #include <algorithm>
 8 #include <string>
 9 #include <queue>
10 #include <vector>
11 using namespace std;
12 const int maxn= 1e5+10;
13 const double eps= 1e-6;
14 const int inf = 0x3f3f3f3f;
15 const int mod =10;
16 typedef long long ll;
17 int n,m;
18 struct matrix
19 {
20     int m[35][35];
21     matrix()
22     {
23         memset(m,0,sizeof(m));
24     }
25 };
26 matrix operator *(const matrix &a,const matrix &b)
27 {
28     matrix c;
29     for(int i=1;i<=n;i++)
30         for(int j=1;j<=n;j++)
31         {
32             c.m[i][j]=0;
33             for(int k=1;k<=n;k++)
34                 c.m[i][j]=(c.m[i][j]+a.m[i][k]*b.m[k][j])%mod;
35         }
36         return c;
37 }
38 matrix quick(matrix base,int pow)
39 {
40     matrix a;
41     for(int i=1;i<=n;i++) a.m[i][i]=1;
42     while(pow)
43     {
44         if(pow&1) a=a*base;
45         base=base*base;
46         pow>>=1;
47     }
48     return a;
49 }
50 matrix sum(matrix a,matrix b)
51 {
52     for(int j=1;j<=n;j++)
53         for(int k=1;k<=n;k++)
54             a.m[j][k]=(a.m[j][k]+b.m[j][k])%mod;
55     return a;
56 }
57 matrix solve(matrix x,int y)
58 {
59     matrix a,s;
60     if(y==1)
61         return x;
62     a=solve(x,y/2);
63     s=sum(a,a*quick(x,y/2));
64     if(y&1)
65         s=sum(s,quick(x,y));
66     return s;
67 }
68 int main()
69 {
70     int t;
71     scanf("%d",&t);
72     while(t--)
73     {
74         scanf("%d %d",&n,&m);
75         matrix  a;
76         for(int i=1;i<=n;i++)
77             for(int j=1;j<=n;j++)
78                 scanf("%d",&a.m[i][j]);
79         matrix ans=solve(a,m);
80         for(int i=1;i<=n;i++)
81         {
82             for(int j=1;j<=n;j++)
83             {
84                 if(j==n)
85                     printf("%d\n",ans.m[i][j]);
86                 else
87                     printf("%d ",ans.m[i][j]);
88             }
89         }
90 
91     }
92 }