HDU

A Simple Math Problem

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4283 Accepted Submission(s): 2573

Problem Description

Lele now is thinking about a simple function f(x).

If x < 10 f(x) = x.
If x >= 10 f(x) = a0 * f(x-1) + a1 * f(x-2) + a2 * f(x-3) + …… + a9 * f(x-10);
And ai(0<=i<=9) can only be 0 or 1 .

Now, I will give a0 ~ a9 and two positive integers k and m ,and could you help Lele to caculate f(k)%m.

Input

The problem contains mutiple test cases.Please process to the end of file.
In each case, there will be two lines.
In the first line , there are two positive integers k and m. ( k<2*10^9 , m < 10^5 )
In the second line , there are ten integers represent a0 ~ a9.

Output

For each case, output f(k) % m in one line.

Sample Input

9999
1 1 1 1 1 1 1 1 1
500
0 1 0 1 0 1 0 1 0

Sample Output

45
104

Author

linle

Source

2007省赛集训队练习赛（6）_linle专场

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这道题算是矩阵算法的一个模板题了，构造矩阵，运用快速幂求解。

#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
typedef long long LL;
typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MOD = 1e9 + 7;
const int MAXN = 100 + 10;
int n, mod, m;
struct Matrix {
    LL m[MAXN][MAXN];
    int row, col;
};
Matrix ori, res, u;
void init() {
    n = 10;
    memset(res.m, 0, sizeof(res.m));
    ori.row = ori.col = n;
}
void scan_in() {
    for (int i = 1; i <= n; i++) {
        scanf("%d", &ori.m[1][i]);
    }
    for(int i = 1; i <= n; i++) {
        ori.m[i][i-1] =1 ;
    }
    u.row = n; u.col = 1;
    for (int i = 1; i <= n; i++) {
        u.m[n - i + 1][1] = i - 1;
    }
}
Matrix multi(Matrix x, Matrix y) {
    Matrix z;
    memset(z.m, 0, sizeof(z.m));
    z.row = x.row; z.col = y.col;
    for (int i = 1; i <= x.row; i++) {
        for (int k = 1; k <= x.col; k++) {
            for (int j = 1; j <= y.col; j++) {
                z.m[i][j] += x.m[i][k]*y.m[k][j]%mod;
            }
            z.m[i][k] %= mod;
        }
    }
    return z;
}
Matrix pow_mod(Matrix a, int x){
    Matrix b;
    b.col = a.col; b.row = a.row;
    memset(b.m, 0, sizeof(b.m));
    for (int i = 1; i <= n; i++) {
        b.m[i][i] = 1;
    }
    while(x){
        if(x&1) b = multi(a,b);
        a = multi(a, a);
        x >>= 1;
    }
    return b;
}
int main() {
    while (scanf("%d%d", &m, &mod) != EOF) {
        init(); scan_in(); int ans = 0;
        if (m < 10) {printf("%d
", m%mod); continue;}
        res = pow_mod(ori, m - 9);
        for(int i = 1;i <= 10; i++) {
            ans += (res.m[1][i]*(10 - i))%mod;
        }
        printf("%d
", ans%mod);
    }
    return 0;
}