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Consider an n-by-n matrix A. We define Ak = A * A * ... * A (k times). Here, * denotes the usual matrix multiplication.
You are to write a program that computes the matrix A + A2 + A3 + ... + Ak.
Suppose A = 
. Then A2 = = 
, thus:
Such computation has various applications. For instance, the above example actually counts all the paths in the following graph:
Input consists of no more than 20 test cases. The first line for each case contains two positive integers n (≤ 40) and k (≤ 1000000). This is followed by n lines, each containing n non-negative integers, giving the matrix A.
Input is terminated by a case where n = 0. This case need NOT be processed.
For each case, your program should compute the matrix A + A2 + A3 + ... + Ak. Since the values may be very large, you only need to print their last digit. Print a blank line after each case.
3 2 0 2 0 0 0 2 0 0 0 0 0
0 2 4 0 0 2 0 0 0
注意到K为偶数时有 A1 + A^2 + A^3 +.... A^k = (E + A^(k /2) (A ^ 1 + A ^ 2 + A ^ 3 + ...A ^ k / 2)
到K为奇数时把最后一想A^k单独提出剩下的为偶数
DFS即可
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cstdlib>
using namespace std;
#define MAXN 50
int n,K;
struct node
{
int mat[MAXN][MAXN];
};
node calcu(node x, node y)
{
node ret;
memset(ret.mat,0,sizeof(ret.mat));
for (int i = 1; i <= n ; i++)
for (int j = 1; j <= n ; j++)
{
for (int k = 1; k <= n ; k++)
ret.mat[i][j] = (ret.mat[i][j] + x.mat[i][k] * y.mat[k][j]) % 10;
}
return ret;
}
node add(node x,node y)
{
node ret;
memset(ret.mat,0,sizeof(ret.mat));
for (int i = 1; i <= n ; i++)
for (int j = 1; j <= n ; j++)
{
ret.mat[i][j] = x.mat[i][j] + y.mat[i][j];
ret.mat[i][j] %= 10;
}
return ret;
}
node pow_mat(node x,int cnt)
{
node ret;
memset(ret.mat,0,sizeof(ret.mat));
for (int i = 1; i < MAXN ; i++) ret.mat[i][i] = 1;
while (cnt)
{
if (cnt & 1) ret = calcu(ret,x);
x = calcu(x,x);
cnt >>= 1;
}
return ret;
}
node dfs(node cur, int k)
{
if (k == 1) return cur;
node res = dfs(cur,k / 2);
node ans;
ans = add(res,calcu(res,pow_mat(cur,k / 2)));
if (k & 1) ans = add(ans,pow_mat(cur,k));
return ans;
}
int main()
{
while (scanf("%d%d",&n,&K) != EOF)
{
if (n == 0) break;
node ans;
for (int i = 1; i <= n ; i++)
for (int j = 1; j <= n ; j++) {scanf("%d",&ans.mat[i][j]); ans.mat[i][j] %= 10;}
node ret = dfs(ans,K);
for (int i = 1; i <= n ; i++)
{
printf("%d",ret.mat[i][1]);
for (int j = 2; j <= n ; j++)
printf(" %d",ret.mat[i][j]);
putchar(‘\n‘);
}
putchar(‘\n‘);
}
return 0;
}
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原文地址：http://www.cnblogs.com/Commence/p/4337158.html
