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  • 1142

    1142 - Summing up Powers (II)
    Time Limit: 2 second(s) Memory Limit: 32 MB

    Shanto is learning how to power up numbers and he found an efficient way to find kth power of a matrix. He was quite happy with his discovery. Suddenly his sister Natasha came to him and asked him to find the summation of the powers. To be specific his sister gave the following problem.

    Let A be an n x n matrix. 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.

    Shanto smiled and thought that it would be an easy one. But after a while he found that it's tough for him. Can you help him?

    Input

    Input starts with an integer T (≤ 20), denoting the number of test cases.

    Each case starts with two integers n (1 ≤ n ≤ 30) and k (1 ≤ k ≤ 109). Each of the next n lines will contain n non-negative integers (not greater than 10).

    Output

    For each case, print the case number and the result matrix. For each cell, just print the last digit. See the samples for more details.

    Sample Input

    Output for Sample Input

    2

    3 2

    1 4 6

    6 5 2

    1 2 3

    3 10

    1 4 6

    6 5 2

    1 2 3

    Case 1:

    208

    484

    722

    Case 2:

    868

    620

    546


    PROBLEM SETTER: JANE ALAM JAN
      1 #include<stdio.h>
      2 #include<algorithm>
      3 #include<iostream>
      4 #include<stdlib.h>
      5 #include<string.h>
      6 #include<math.h>
      7 #include<queue>
      8 using namespace std;
      9 typedef long long LL;
     10 typedef struct node
     11 {
     12     int  m[70][70];
     13     node()
     14     {
     15         memset(m,0,sizeof(m));
     16     }
     17 } maxtr;
     18 int  ans[70][70];
     19 void E(node *nn,int n);
     20 maxtr ju(int n);
     21 maxtr quick(node ju,int n,int m);
     22 int main(void)
     23 {
     24     int i,j,k;
     25     int n,m;
     26     int s;
     27     cin>>k;
     28     for(s=1; s<=k; s++)
     29     {
     30         scanf("%d %d",&n,&m);
     31         for(i=0; i<n; i++)
     32         {
     33             for(j=0; j<n; j++)
     34             {
     35                 scanf("%d",&ans[i][j]);
     36                 ans[i][j]%=10;
     37             }
     38         }node aa;aa=ju(n);
     39         aa=quick(aa,n,m);printf("Case %d:
    ",s);
     40         for(i=0;i<n;i++)
     41         {
     42             for(j=n;j<2*n;j++)
     43             {
     44                 printf("%d",aa.m[i][j]);
     45             }printf("
    ");
     46         }
     47     }return 0;
     48 }
     49 void E(node *nn,int n)
     50 {
     51     int i,j,k;
     52     for(i=0; i<n; i++)
     53     {
     54         for(j=0; j<n; j++)
     55         {
     56             if(i==j)
     57                 nn->m[i][j]=1;
     58             else nn->m[i][j]=0;
     59         }
     60     }
     61 }
     62 maxtr ju(int n)
     63 {
     64     int i,j,k;
     65     maxtr nn;
     66     for(i=0; i<n; i++)
     67     {
     68         for(j=0; j<n; j++)
     69         {
     70             nn.m[i][j]=ans[i][j];
     71         }
     72     }
     73     for(i=0; i<n; i++)
     74     {
     75         for(j=n; j<2*n; j++)
     76         {
     77             nn.m[i][j]=ans[i][j-n];
     78         }
     79     }
     80     node cc;
     81     E(&cc,n);
     82     for(i=n; i<2*n; i++)
     83     {
     84         for(j=n; j<2*n; j++)
     85         {
     86             nn.m[i][j]=cc.m[i-n][j-n];
     87         }
     88     }return nn;
     89 }
     90 maxtr quick(node ju,int n,int m)
     91 {   node ee;
     92 
     93     E(&ee,2*n);
     94     int i,j,k;
     95     int s;
     96     while(m)
     97     {
     98         if(m&1)
     99         {
    100             node cc;
    101             for(i=0; i<2*n; i++)
    102             {
    103                 for(j=0; j<2*n; j++)
    104                 {
    105                     for(s=0; s<2*n; s++)
    106                     {
    107                         cc.m[i][j]=(ju.m[i][s]*ee.m[s][j]+cc.m[i][j])%10;
    108                     }
    109                 }
    110             }
    111             ee=cc;
    112         }
    113         node cc;
    114         for(i=0; i<2*n; i++)
    115         {
    116             for(j=0; j<2*n; j++)
    117             {
    118                 for(s=0; s<2*n; s++)
    119                 {
    120                     cc.m[i][j]=(ju.m[i][s]*ju.m[s][j]+cc.m[i][j])%10;
    121                 }
    122             }
    123         }
    124         ju=cc;
    125         m/=2;
    126     }
    127     return ee;
    128 }
    油!油!you@
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  • 原文地址:https://www.cnblogs.com/zzuli2sjy/p/5494718.html
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