hdoj 5015 233 Matrix(矩阵的构造)

一开始根本没想到可以这样构造矩阵,看了半天才看懂,太巧了这题。。


较好的讲解:点击打开链接


还要注意的是要使用long long


代码:

#include
#include
#include
using namespace std;
const int mod = 10000007;
int n, m, N;
struct node
{
    long long s[15][15];
};

node mul(node a, node b)
{
    node t;
    memset(t.s, 0, sizeof(t.s));
    for(int i = 0; i < N; i++)
        for(int j = 0; j < N; j++)
            for(int k = 0; k < N; k++)
                t.s[i][j] = (t.s[i][j]+a.s[i][k]*b.s[k][j])%mod;
    return t;
}

node mt_pow(node p, int k)
{
    node q;
    memset(q.s, 0, sizeof(q.s));
    for(int i = 0; i < N; i++)
        q.s[i][i] = 1;
    while(k)
    {
        if(k&1) q = mul(p, q);
        p = mul(p, p);
        k /= 2;
    }
    return q;
}

int main(void)
{
    int a[15];
    while(~scanf("%d%d", &n, &m))
    {
        N = n+2;
        a[0] = 23, a[n+1] = 3;
        for(int i = 1; i <= n; i++)
            scanf("%d", &a[i]);
        node base;
        //init
        memset(base.s, 0, sizeof(base.s));
        for(int i = 0; i <= n; i++)
            base.s[i][0] = 10, base.s[i][n+1] = 1;
        base.s[n+1][n+1] = 1;
        for(int i = 1; i <= n; i++)
            for(int j = 1; j <= i; j++)
                base.s[i][j] = 1;
        node ans = mt_pow(base, m);
        int res = 0;
        for(int i = 0; i < N; i++)
            res = (res+a[i]*ans.s[n][i])%mod;
        printf("%d\n", res);
    }
    return 0;
}


233 Matrix

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 2031    Accepted Submission(s): 1185


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
 
   
1 1 1 2 2 0 0 3 7 23 47 16
 


Sample Output
 
   
234 2799 72937
Hint
 


Source
2014 ACM/ICPC Asia Regional Xi'an Online
 




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