uva11149

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 . Example Suppose A =   0 2 0 0 0 2 0 0 0  . Then A2 =   0 2 0 0 0 2 0 0 0     0 2 0 0 0 2 0 0 0   =   0 0 4 0 0 0 0 0 0  , thus: A + A2 =   0 2 0 0 0 2 0 0 0   +   0 0 4 0 0 2 0 0 0   =   0 2 4 0 0 2 0 0 0   Such computation has various applications. For instance, the above example actually counts all the paths in the following graph: Input 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. Output 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. Sample Input 3 2 0 2 0 0 0 2 0 0 0 0 0 Sample Output 0 2 4 0 0 2 0 0 0

矩阵快速幂+分治。。。

很巧妙啊 先开始还在想怎么错位相减。。。

具体细节不讲了 代码里都有 这道题给我们的启示是碰见这种连续幂相加的东西要想分治。。。

先开始t了半天，结果写成暴力了。。。

#include<bits/stdc++.h>
using namespace std;
const int N = 41;
struct mat {
    int a[N][N];
} A;
int n, k;
mat operator * (mat A, mat B)
{
    mat ret; memset(ret.a, 0, sizeof(ret.a));
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j)
             for(int k = 1; k <= n; ++k) ret.a[i][j] = (ret.a[i][j] + A.a[i][k] % 10 * B.a[k][j] % 10) % 10;
    return ret;         
}
mat power(mat x, int t)
{
    mat ret; memset(ret.a, 0, sizeof(ret.a));
    for(int i = 1; i <= n; ++i) ret.a[i][i] = 1;
    for(; t; t >>= 1, x = x * x) if(t & 1) ret = ret * x;
    return ret;
}
mat operator + (mat A, mat B)
{
    for(int i = 1; i <= n; ++i)
        for(int j = 1; j <= n; ++j) A.a[i][j] = (A.a[i][j] + B.a[i][j]) % 10;
    return A;    
}
mat solve(int t)
{
    if(t == 1) return A;
    mat x = solve(t / 2), ret = x, B = power(A, t / 2);
    if(t & 1) return x + (x + B * A) * B;
    else return x + x * B; 
}
int main()
{
    while(scanf("%d%d", &n, &k))
    {
        if(n == 0) break;
        for(int i = 1; i <= n; ++i) 
            for(int j = 1; j <= n; ++j) scanf("%d", &A.a[i][j]), A.a[i][j] %= 10;
        mat x = solve(k);     
        for(int i = 1; i <= n; ++i)
        {
            for(int j = 1; j < n; ++j) printf("%d ", x.a[i][j]);
            printf("%d
", x.a[i][n]);
        }
        puts("");
    }
    return 0;
}