ACM学习历程——HDU5015 233 Matrix（矩阵快速幂）（2014陕西网赛）

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <string>
#define inf 0x3fffffff
#define esp 1e-10
#define N 10000007
#define LL long long

using namespace std;

struct Mat
{
    LL val[15][15];
    int len;

    Mat operator = (const Mat& a)
    {
        for (int i = 0; i < len; ++i)
            for (int j = 0; j < len; ++j)
                val[i][j] = a.val[i][j];
        len = a.len;
        return *this;
    }

    Mat operator * (const Mat& a)
    {
        Mat x;
        memset(x.val, 0, sizeof(x.val));
        x.len = len;
        for (int i = 0; i < len; ++i)
            for (int j = 0; j < len; ++j)
                for (int k = 0; k < len; ++k)
                    if (val[i][k] && a.val[k][j])
                        x.val[i][j] = (x.val[i][j] + (val[i][k]*a.val[k][j])%N)%N;
        return x;
    }

    Mat operator ^ (const int& a)
    {
        int n = a;
        Mat x, p = *this;
        memset(x.val, 0, sizeof(x.val));
        x.len = len;
        for (int i = 0; i < len; ++i)
            x.val[i][i] = 1;
        while (n)
        {
            if (n & 1)
                x = x * p;
            p = p * p;
            n >>= 1;
        }
        return x;
    }
};

int n, m;
LL a[15], ans;

void Make(Mat &p)
{
    p.len = n + 2;
    memset(p.val, 0, sizeof(p.val));
    for (int i = 0; i <= n; ++i)
        p.val[i][0] = 10;
    for (int i = 0; i <= n+1; ++i)
        p.val[i][n+1] = 1;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= i; ++j)
            p.val[i][j] = 1;
}

int main()
{
    //freopen("test.txt", "r", stdin);
    while (scanf("%d%d", &n, &m) != EOF)
    {
        Mat p;
        Make(p);
        p = p ^ m;
        a[0] = 23;
        a[n+1] = 3;
        for (int i = 1; i <= n; ++i)
            scanf("%I64d", &a[i]);
        ans = 0;
        for (int i = 0; i <= n+1; ++i)
            ans = (ans + (p.val[n][i]*a[i])%N)%N;
        printf("%I64d\n", ans);
    }
    return 0;
}