HDU 5950Recursive sequence ICPC沈阳站

Recursive sequence

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1525 Accepted Submission(s): 710

Problem Description

Farmer John likes to play mathematics games with his N cows. Recently, they are attracted by recursive sequences. In each turn, the cows would stand in a line, while John writes two positive numbers a and b on a blackboard. And then, the cows would say their identity number one by one. The first cow says the first number a and the second says the second number b. After that, the i-th cow says the sum of twice the (i-2)-th number, the (i-1)-th number, and

Input

The first line of input contains an integer t, the number of test cases. t test cases follow.
Each case contains only one line with three numbers N, a and b where N,a,b <

Output

For each test case, output the number of the N-th cow. This number might be very large, so you need to output it modulo 2147493647.

Sample Input

3 1 2

4 1 10

Sample Output

369

Hint

In the first case, the third number is 85 = 2*1十2十3^4.

In the second case, the third number is 93 = 2*1十1*10十3^4 and the fourth number is 369 = 2 * 10 十 93 十 4^4.

递推超时，矩阵快速幂

#pragma comment(linker, "/STACK:102400000,102400000")
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <cstdlib>
#include <map> 
#include <set>
#include <ctime> 
#include <queue> 

#define LL long long

using namespace std;

const LL _MOD = 2147493647, maxN = 4, MOD = _MOD*2;

int n;

LL f(int _n)
{
    LL n = _n, ans =200, t=1;
    t = t*n%MOD; ans = (ans + t*139)%MOD;
    t = t*n%MOD; ans = (ans + t*48)%MOD;
    t = t*n%MOD; ans = (ans + t*10)%MOD;
    t = t*n%MOD; ans = (ans + t)%MOD;
    return ans/2 % _MOD;
}

struct matrix
{
    int n, m;
    LL a[maxN][maxN];
    LL* operator [](int x) {return a[x];}
    void print()
    {
        for(int i = 1; i <= n; i++)
        {
            for(int j = 1; j <= m; j++)
                printf("%d ", a[i][j]);
            printf("
");
        }
        printf("
");
    }
};

matrix operator *(matrix a, matrix b)
{
    matrix c; c.n = a.n; c.m = b.m;
    memset(c.a, 0, sizeof(c.a));
    LL tmp;
    for(int i = 1; i <= a.n; i++)
    {
        tmp = 0;
        for(int j = 1; j <= b.m; j++)
        {
            for(int k = 1; k <= a.m; k++)    tmp = (tmp+a[i][k] * b[k][j])%_MOD;
            c[i][j] = tmp % _MOD;
            tmp = 0;
        }
    }
    return c;
}

matrix operator ^(matrix a, LL x)
{
    matrix b;
    memset(b.a, 0, sizeof(b.a));
    b.n = a.n; b.m = a.m;
    for(int i=1; i <= a.n; i++)    b[i][i]=1;
    for(;x;a=a*a,x>>=1)    if(x&1)    b=b*a;
    return b;
}

int main()
{
//    cout<<2*f(3)+f(4)-f(5)<<endl;
//    return 0;
    #ifndef ONLINE_JUDGE
    freopen("test_in.txt", "r", stdin); 
    //freopen("test_out.txt", "w", stdout);
    #endif
    int T; scanf("%d", &T);
    while(T--)
    {
        int a, b, n; scanf("%d%d%d", &n, &a, &b);
        LL _a = a; _a += f(1); LL _b = b; _b += f(2);
        matrix m; m.n = m.m = 2; m[1][1] = _a; m[1][2] = _b; m[2][1] = m[2][2] = 0;
        matrix t; t.n = t.m = 2; t[1][1] = 0; t[1][2] = 2;  t[2][1] = t[2][2] = 1;
        t = t^(n-2); 
        m = m*t;
        LL ans = (m[1][2] - f(n) + _MOD) % _MOD;
        printf("%d
", (int)ans);
    }
}