hdu

题意：当x < 10时， f(x) = x；

当x >= 10 时，f(x) = a0 * f(x-1) + a1 * f(x-2) +  + a2 * f(x-3) + …… + a9 * f(x-10)；
ai(0<=i<=9) 只能是0或者1 ，给出a0 ~ a9，k和m，计算f(k)%m（k<2*10^9 , m < 10^5）。

题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=1757

——>>构造矩阵，快速幂求解。

用excel弄了个~~

#include <cstdio>
#include <cstring>

using namespace std;

const int maxn = 10 + 5;
int k, mod, n;

struct Mar{     //矩阵
    int m[maxn][maxn];
    Mar(){
        memset(m, 0, sizeof(m));
    }
};

Mar operator + (Mar a, Mar b){      //矩阵+
    Mar ret;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++) ret.m[i][j] = (a.m[i][j] + b.m[i][j]) % mod;
    return ret;
}

Mar operator * (Mar a, Mar b){      //矩阵*
    Mar ret;
    for(int i = 0; i < n; i++)
        for(int j = 0; j < n; j++)
            for(int l = 0; l < n; l++) ret.m[i][j] = (ret.m[i][j] + a.m[i][l] * b.m[l][j]) % mod;
    return ret;
}

Mar pow_mod(Mar a, int n){      //矩阵快速幂
    if(n == 1) return a;
    Mar x = pow_mod(a, n/2);
    x = x * x;
    if(n&1) x = x * a;
    return x;
}

void solve(){
    Mar A;
    n = 10;
    int i, j, ret = 0;
    for(i = 0; i < n; i++) scanf("%d", &A.m[0][i]);
    if(k < 10) ret = k % mod;
    else{
        for(i = 1, j = 0; i < n; i++, j++) A.m[i][j] = 1;
        A = pow_mod(A, k-9);
        for(i = 0, j = 9; i < n; i++, j--) ret = (ret + A.m[0][i] * j) % mod;
    }
    printf("%d
", ret);
}

int main()
{
    while(scanf("%d%d", &k, &mod) == 2) solve();
    return 0;
}