Hdu6050 Funny Function（2017多校第2场）

Funny Function

                                                                         Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
                                                                                                      Total Submission(s): 422    Accepted Submission(s): 180

Problem Description

Function  Fx,y satisfies:

For given integers N and M,calculate  Fm,1  modulo 1e9+7.

 

Input

There is one integer T in the first line.
The next T lines,each line includes two integers N and M .
1<=T<=10000,1<=N,M<2^63.

 

Output

For each given N and M,print the answer in a single line.

 

Sample Input

  
  

   
   2
2 2
3 3
  
  

 

Sample Output

  
  

   
   2
33
  
  

 

Source

2017 Multi-University Training Contest - Team 2

 

Recommend

liuyiding

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题目的意思是根据给出的式子，求第m行第一个是多少

思路：nm巨大绝对找规律或（矩阵）快速幂，现场找了1个多小时规律（还是太菜），打表找规律发现：

n为偶数是：am=2*(2^n -1)^(m-1)/3;

n为奇数是:am=(2^n-1)am-1-c(n/2)   c(x)=4*c(x-1)+2;

矩阵快速幂搞一波。

注意偶数时除法取模要逆元

#include <iostream>  
#include <cstdio>  
#include <cstring>  
#include <string>  
#include <algorithm>  
#include <map>  
#include <cmath>
#include <set>  
#include <stack>  
#include <queue>  
#include <vector>  
#include <bitset>  
#include <functional>

using namespace std;

#define LL long long  
const int INF = 0x3f3f3f3f;
const LL mod = 1000000007;

LL n, m;

struct Matrix
{
    LL v[9][9];
    Matrix()
    {
        memset(v, 0, sizeof v);
    }
} dan;

Matrix mul(Matrix a, Matrix b, int d)
{
    Matrix ans;
    for (int i = 0; i < d; i++)
    {
        for (int j = 0; j < d; j++)
        {
            for (int k = 0; k < d; k++)
            {
                ans.v[i][j] += (a.v[i][k] * b.v[k][j]) % mod;
                ans.v[i][j] %= mod;
            }
        }
    }
    return ans;
}

Matrix qpow(Matrix a, LL k, int d)
{
    Matrix ans = dan;
    while (k)
    {
        if (k & 1) ans = mul(ans, a, d);
        k >>= 1;
        a = mul(a, a, d);
    }
    return ans;
}

LL qpow(LL x, LL y)
{
    LL ans = 1;
    while (y)
    {
        if (y & 1) ans *= x, ans %= mod;
        y >>= 1;
        x *= x;
        x %= mod;
    }
    return ans;
}

LL extend_gcd(LL a, LL b, LL &x, LL &y)
{
    if (!b)
    {
        x = 1, y = 0;
        return a;
    }
    LL gcd = extend_gcd(b, a%b, x, y);
    LL tmp = x;
    x = y;
    y = tmp - (a / b)*y;
    return gcd;
}

int main()
{
    int t;
    scanf("%d", &t);
    while (t--)
    {
        scanf("%lld%lld", &n, &m);
        if (m == 1) { printf("1\n"); continue; }
        Matrix a, ans;
        LL ans1 = (qpow(2, n) - 1 + mod) % mod;
        if (n % 2)
        {
            dan.v[0][0] = 0, dan.v[0][1] = 1;
            a.v[0][0] = 4, a.v[0][1] = 0, a.v[1][0] = 2, a.v[1][1] = 1;
            ans = qpow(a, n / 2, 2);
            dan.v[0][0] = 1, dan.v[0][1] = 1;
            a.v[0][0] = ans1, a.v[0][1] = 0, a.v[1][0] = -ans.v[0][0], a.v[1][1] = 1;
            ans = qpow(a, m - 1, 2);
            printf("%lld\n", (ans.v[0][0]+mod)%mod);
        }
        else
        {
            ans1 = qpow(ans1, m - 1);
            LL x, y;
            extend_gcd(3, mod, x, y);
            ans1 = ans1*x%mod;
            ans1 = (ans1 * 2) % mod;
            printf("%lld\n", ans1);
        }
    }
    return 0;
}