C-Recursive Sequence【矩阵快速幂·暑假训练赛4】

最新推荐文章于 2024-10-30 23:46:31 发布

葵五_

最新推荐文章于 2024-10-30 23:46:31 发布

阅读量183

点赞数

CC 4.0 BY-SA版权

分类专栏： # 矩阵快速幂

本文链接：https://blog.youkuaiyun.com/qq_43611144/article/details/98997166

矩阵快速幂专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种递归数列问题，农民John和他的奶牛们玩数学游戏，涉及递归序列的计算。文章提供了问题描述，输入输出样例，并深入探讨了矩阵快速幂的解决方案，包括关键步骤和常见错误，以及正确的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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 i^4. Now, you need to write a program to calculate the number of the N-th cow in order to check if John’s cows can make it right.

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 < 2^31 as described above.

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.

Hint In the first case, the third number is 85 = 2 ∗ 1 + 2 + 3^4.
In the second case, the third number is 93 = 2 ∗ 1 + 1 ∗ 10 + 3^4 and the fourth number is 369 =2 ∗ 10 + 93 + 4^4.

Sample Input

2
3 1 2
4 1 10

Sample Output

85
369

Code

关于矩阵快速幂的讲解，附上网页链接：https://www.cnblogs.com/cmmdc/p/6936196.html

矩阵快速幂的精髓，概括起来应该是两点：
一、模板；
二、找出联系n与n-1的关系矩阵

本题正解–https://blog.youkuaiyun.com/westbrook1998/article/details/82291445

本人两个回合的错误经历↓

//round 1
//Runtime Error, 栈溢出
#include <iostream>
using namespace std;
typedef long long ll;
ll a,b;

ll f(int n){
    if(n==1) return a;
    if(n==2) return b;
    ll ans;
    ans=2*f(n-2)+f(n-1)+n*n*n*n;
    return ans;
}

int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n>>a>>b;
        cout<<(f(n)%2147493647)<<endl;
    }
    return 0;
}

//round 2
//Time Limit Exceeded
#include <iostream>
#include <cmath>
using namespace std;
typedef long long ll;

int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        ll a,b;
        cin>>n>>a>>b;
        if(n==1){
            cout<<(a%2147493647)<<endl;
            continue;
        }
        if(n==2){
            cout<<(b%2147493647)<<endl;
            continue;
        }
        ll n1=a,n2=b,ans;
        int m=3;
        while(true){
            ans=2*n1+n2+pow(m,4);
            if(m==n){
                cout<<(ans%2147493647)<<endl;
                break;
            }
            n1=n2;
            n2=ans;
            m++;
        }
    }
    return 0;
}