A + B Problem + 数论

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本文介绍了一个基于Fibonacci数列的加法问题，通过特定的数学技巧实现两个由Fibonacci数表示的整数相加，并提供了解决方案的详细步骤及示例。

A + B Problem

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 458 Accepted Submission(s): 153

Problem Description

Do you know the famous Fibonacci sequence? It is defined by the recurrence
F₀ = 0, F₁ = 1 and F_n = F_n-1 + F_n-2 for n ≥ 2.
The Fibonacci numbers have many interesting properties. One of them is that the Fibonacci numbers can be used to represent integers. Every positive integer has a unique representation of the form
n = F_k1 + F_k2 + … + F_km, k_i ≥ k_i-1 + 2 for 2 ≤ i ≤ m and k₁ ≥ 2
For example, 6 can be represented as F₂+F₅ and 12 can be represented as F₂+F₄+F₆.
Now you know how to represent positive integers with the Fibonacci numbers, can you add them? Given two Fibonacci formed integers, you should calculate the sum of them.

Input

The first line contains a single integer T, indicating the number of test cases.
Each test case contains two lines; each line contains a single integer m followed by m integers k₁, k₂, …, k_m indicate a Fibonacci formed integer F_k1+F_k2+…+F_km.
The input will be always correct.

Technical Specification

1. 1 <= T <= 100
2. 1 <= m <= 100
3. 2 <= k_i <= 1000000

Output

For each test case, output the case number first, then a single line indicates the sum of the two Fibonacci formed integer. The sum should be Fibonacci formed like the input.

Sample Input

Sample Output

Case 1:
1 5
Case 2:
2 5 7

算法：

主要是调用两个公式 f[n]=f[n-1]+f[n-2]

下一个公式 2*f[n]=f[n-2]+f[n+1]

这两个公式进行求解要注意的是输出结果序号的最小间隔为2

要注意的是 2 3 5

2 3 5

这种特殊的情况出现了序号 1 得出f[1]=f[2]的结果。

#include<stdio.h>
#include<string.h>
#define MAX 1000000
int a[1000010];
int b[1000];
int main()
{
	int x,n,i,j,num,max=0,sum,ji=0;
	scanf("%d",&x);
	while(x--)    
	{
		ji++;
		memset( a, 0, sizeof(a) );  //输入数据
		for( i=1; i<=2; i++ )         
		{
			scanf("%d",&n);
			for( j=1; j<=n;j++ )          
			{
				scanf("%d",&num);
				if( num>max )max=num;
				a[num]++;
			}  
		}
		
		
		for( i=max; i>=2; i--)  //对含有两个重复利用 2*f[n]=f[n+1]+f[n-2]
		{
			if(a[i]>=2){
				a[i]-=2;        
				a[i+1]+=1; 
				a[i-2]+=1;
			}
		} 
		
		if(a[1]!=0)a[2]+=a[1];     //考虑 fibnaci数列中 序号 f[1]=f[2]这种特殊情况
		
		for( i=max+1; i>=3; i-- )     //连续两次调用f[n]=f[n-1]+f[n-2]
		{
			if(a[i]==1&&a[i-1]==1)
			{
				a[i]=a[i-1]=0;
				a[i+1]+=1;  ;
			}             
		}
		
		
		for( i=max+2; i>=3; i-- )
		{
			if(a[i]==1&&a[i-1]==1)
			{
				a[i]=a[i-1]=0;
				a[i+1]+=1;  ;
			}             
		}
		
		
		j=0;
		sum=0;
		for(i=2; i<=max+5; i++)   //输出结果
		{
			if(a[i]!=0){b[j++]=i;sum++;}         
		}
		printf("Case %d:\n",ji);
		printf("%d",sum);
		for( i=0; i<j; i++)
			printf(" %d",b[i]);
		printf("\n");               
	} 
	return 0;
}