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  • HDU 5015 233 Matrix(网络赛1009) 矩阵快速幂

    先贴四份矩阵快速幂的模板:http://www.cnblogs.com/shangyu/p/3620803.html

    http://www.cppblog.com/acronix/archive/2010/08/23/124470.aspx?opt=admin

    http://www.cnblogs.com/vongang/archive/2012/04/01/2429015.html

    http://www.cnblogs.com/yan-boy/archive/2012/11/29/2795294.html

    233 Matrix

    Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others) Total Submission(s): 257    Accepted Submission(s): 165

    Problem Description
       In our daily life we often use 233 to express our feelings. Actually, we may say 2333, 23333, or 233333 ... in the same meaning. And here is the question: Suppose we have a matrix called 233 matrix. In the first line, it would be 233, 2333, 23333... (it means a0,1 = 233,a0,2 = 2333,a0,3 = 23333...) Besides, in 233 matrix, we got ai,j = ai-1,j +ai,j-1( i,j ≠ 0). Now you have known a1,0,a2,0,...,an,0, could you tell me an,m in the 233 matrix?
     
    Input
       There are multiple test cases. Please process till EOF.
       For each case, the first line contains two postive integers n,m(n ≤ 10,m ≤ 109). The second line contains n integers, a1,0,a2,0,...,an,0(0 ≤ ai,0 < 231).
     
    Output
       For each case, output an,m mod 10000007.
     
    Sample Input
    1 1 1 2 2 0 0 3 7 23 47 16
     
    Sample Output
    234 2799 72937
    Hint
     
    Source
     
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     题解1:http://www.cnblogs.com/whatbeg/p/3971994.html

    题解2:http://blog.csdn.net/u013368721/article/details/39271565

    题目分析:矩阵快速幂,构建一个如下的矩阵即可:

    1. n+2行的矩阵  
    2. --                      --   --  --  
    3. | 1  1  1  1  1  1  1  0 |   | a1 |  
    4. | 0  1  1  1  1  1  1  0 |   | a2 |  
    5. | 0  0  1  1  1  1  1  0 |   | a3 |  
    6. | 0  0  0  1  1  1  1  0 |   | a4 |  
    7. | 0  0  0  0  1  1  1  0 | * | a5 |  
    8. | 0  0  0  0  0  1  1  0 |   | an |  
    9. |  - - - - - - - - - - - |   |    |  
    10. | 0  0  0  0  0  0 10  1 |   | 233|  
    11. | 0  0  0  0  0  0  0  1 |   | 3  |  
    12. --                      --   --  --  
      1 #include<iostream>
      2 #include<cstring>
      3 #include<cstdlib>
      4 #include<cstdio>
      5 #include<algorithm>
      6 #include<cmath>
      7 #include<queue>
      8 #include<map>
      9 #include<string>
     10 
     11 #define N 15
     12 #define M 15
     13 #define mod 10000007
     14 #define p 10000007
     15 #define mod2 100000000
     16 #define ll long long
     17 #define LL long long
     18 #define maxi(a,b) (a)>(b)? (a) : (b)
     19 #define mini(a,b) (a)<(b)? (a) : (b)
     20 
     21 using namespace std;
     22 
     23 ll nn,m;
     24 ll n;
     25 ll x[15];
     26 //ll ans;
     27 
     28 struct Mat
     29 {
     30     ll mat[N][N];
     31 };
     32 
     33 Mat e,f,g;
     34 Mat operator * (Mat a,Mat b)
     35 {
     36     Mat c;
     37     memset(c.mat,0,sizeof(c.mat));
     38     ll i,j,k;
     39     for(k =0 ; k < n ; k++)
     40     {
     41         for(i = 0 ; i < n ;i++)
     42         {
     43             if(a.mat[i][k]==0) continue;//优化
     44             for(j = 0 ;j < n ;j++)
     45             {
     46                 if(b.mat[k][j]==0) continue;//优化
     47                 c.mat[i][j] = (c.mat[i][j]+(a.mat[i][k]*b.mat[k][j])%mod)%mod;
     48             }
     49         }
     50     }
     51     return c;
     52 }
     53 Mat operator ^(Mat a,ll k)
     54 {
     55     Mat c;
     56     ll i,j;
     57     for(i =0 ; i < n ;i++)
     58         for(j = 0; j < n ;j++)
     59         c.mat[i][j] = (i==j);
     60     for(; k ;k >>= 1)
     61     {
     62         if(k&1) c = c*a;
     63         a = a*a;
     64     }
     65     return c;
     66 }
     67 
     68 
     69 void ini()
     70 {
     71     ll i,j;
     72     for(i=1;i<=nn;i++){
     73         scanf("%I64d
    ",&x[i]);
     74     }
     75     memset(e.mat,0,sizeof(e.mat));
     76     memset(f.mat,0,sizeof(f.mat));
     77     e.mat[0][0]=233;
     78     e.mat[0][1]=3;
     79     e.mat[0][2]=233+x[1];
     80     for(i=2;i<=nn;i++){
     81         e.mat[0][i+1]=e.mat[0][i]+x[i];
     82     }
     83     for(j=0;j<nn+2;j++){
     84         if(j!=1){
     85             f.mat[0][j]=10;
     86         }
     87         f.mat[1][j]=1;
     88     }
     89     for(i=2;i<nn+2;i++){
     90         for(j=i;j<nn+2;j++){
     91             f.mat[i][j]=1;
     92         }
     93     }
     94     n=nn+2;
     95 }
     96 
     97 void solve()
     98 {
     99     if(m>1){
    100          g= e*  (f^(m-1) );
    101     }
    102     else{
    103         g.mat[0][nn+1]=e.mat[0][nn+1];
    104     }
    105 }
    106 
    107 void out()
    108 {
    109     printf("%I64d
    ",g.mat[0][nn+1]);
    110 }
    111 
    112 int main()
    113 {
    114    // freopen("data.in","r",stdin);
    115   //  freopen("data.out","w",stdout);
    116     //scanf("%d",&T);
    117     //for(int cnt=1;cnt<=T;cnt++)
    118    // while(T--)
    119     while(scanf("%I64d%I64d",&nn,&m)!=EOF)
    120     {
    121         ini();
    122         solve();
    123         out();
    124     }
    125 
    126     return 0;
    127 }
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  • 原文地址:https://www.cnblogs.com/njczy2010/p/3972316.html
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