hdu5015（矩阵快速幂）

最新推荐文章于 2021-08-05 13:56:54 发布

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最新推荐文章于 2021-08-05 13:56:54 发布

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探讨233Matrix矩阵的构建方式与求解过程，通过矩阵快速幂算法解决特定矩阵中元素值的问题。

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地址：http://acm.hdu.edu.cn/showproblem.php?pid=5015

233 Matrix

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 855 Accepted Submission(s): 514

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 a _0,1 = 233,a _0,2 = 2333,a _0,3 = 23333...) Besides, in 233 matrix, we got a _i,j = a _i-1,j +a _i,j-1( i,j ≠ 0). Now you have known a _1,0,a _2,0,...,a _n,0, could you tell me a _n,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 ≤ 10 ⁹). The second line contains n integers, a _1,0,a _2,0,...,a _n,0(0 ≤ a _i,0 < 2 ³¹).

Output

For each case, output a _n,m mod 10000007.

Sample Input

Sample Output

题意：有一个从（0,0）点开始(n+1)*(m+1)的矩阵，其（0,i）点的格式为233……3（其中有i+1个3，i从1开始），其（i,0）点依次输入进去，然后剩下的点位该点前边的点与上面的点之和，问该矩阵右下角的点的值。

思路：很明显的矩阵快速幂，推出矩阵后就好些了。

代码：

/***************************************************
     写这题的报告主要是为了矩阵快速幂的模板
***************************************************/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define LL __int64
#define Mod 10000007
struct Mat {
    LL mat[15][15];
    void cle(){
        memset(mat,0,sizeof(mat));
    }
}two,three;
int m,n;
Mat operator * (Mat a, Mat b) {  //矩阵快速幂的模板
    Mat c;
    c.cle();  //清空c矩阵
    for(int k = 0; k <= m+1; ++k) {  //注意乘的范围
        for(int i = 0; i <= m+1; ++i) {
            if(a.mat[i][k] <= 0)  continue;
            for(int j = 0; j <= m+1; ++j) {
                if(b.mat[k][j] <= 0)    continue;
                c.mat[i][j] += a.mat[i][k] * b.mat[k][j];
                if(c.mat[i][j]>Mod) c.mat[i][j]%=Mod;  //有取mod的在这里取mod
            }
        }
    }
    return c;
}
int main(){
    while(scanf("%d%d",&m,&n)>0){
        two.cle();
        three.cle();
        two.mat[0][0]=233;
        two.mat[0][m+1]=3;
        for(int i=1;i<=m;i++)
            scanf("%d",&two.mat[0][i]);
        for(int i=0;i<=m;i++)
            for(int j=i;j<=m;j++)
                three.mat[i][j]=1;
        three.mat[0][0]=10;
        three.mat[m+1][0]=1;
        three.mat[m+1][m+1]=1;
        while(n){
            if(1&n){
                two=two*three;
            }
            three=three*three;
            n/=2;
        }
        printf("%I64d\n",two.mat[0][m]);
    }
    return 0;
}