杭电acm--1005

最新推荐文章于 2021-05-19 02:11:33 发布

qq_26919527

最新推荐文章于 2021-05-19 02:11:33 发布

阅读量373

点赞数

分类专栏： ACM

ACM 专栏收录该内容

84 篇文章

订阅专栏

本文介绍了一种高效计算特定数列f(n)的方法，该数列定义为f(1)=1, f(2)=1, f(n)=(A*f(n-1)+B*f(n-2)) mod 7。通过循环而非递归的方式避免了内存溢出问题，并利用周期性特征进一步优化计算过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

A number sequence is defined as follows:

f(1) = 1, f(2) = 1, f(n) = (A * f(n - 1) + B * f(n - 2)) mod 7.

Given A, B, and n, you are to calculate the value of f(n).

Inpu

The input consists of multiple test cases. Each test case contains 3 integers A, B and n on a single line (1 <= A, B <= 1000, 1 <= n <= 100,000,000). Three zeros signal the end of input and this test case is not to be processed.

Output

For each test case, print the value of f(n) on a single line.

#include <iostream>
 using namespace std;
 int main()
 {
  int a,b,n,i,f[53];
  while(cin>>a>>b>>n)
  {
   if(a==0 && b==0 && n==0) break;
   if(n==1 || n==1)
   {
    cout<<"1"<<endl;
    continue;
   }
   f[1]=1,f[2]=1;
   a%=7,b%=7;
   for(i=3;i<=52;i++)
   {
    f[i]=(a*f[i-1]+b*f[i-2])%7;
    if(f[i-1]==1 && f[i]==1) break;
   }
   i=i-2;
   n%=i;
   f[0]=f[i];
   cout<<f[n]<<endl;
  }
  return 0;
 }

下面是我用递归写的，不过Memory Limit Exceeded

#include<stdio.h>
#include<stdlib.h>
#include<iostream>
#include<cmath>
#include<iomanip>

//#define P 3.141592653

using namespace std;

int function(int A, int B, int n)
{
	if (n ==1)
		return 1;
	else if (n == 2)
		return 1;
	else if (n == 3)
		return (A  + B ) % 7;
	else if (n > 3)
		return (A*function(A, B, n - 1)+B*function(A,B,n-2))%7;

}
void main()
{
	int n;
	int A, B;
	while (cin >> A >> B >> n)
	{
		if (A == 0 && B == 0 && n == 0)
			break;

		cout << function(A, B, n) << endl;

	}
	system("pause");
}