BestCoder Round #28

最新推荐文章于 2021-05-16 21:56:47 发布

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最新推荐文章于 2021-05-16 21:56:47 发布

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分类专栏： OJ编程文章标签： poj

本文链接：https://blog.csdn.net/codeforcer/article/details/43369855

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本文深入探讨了两个技术挑战：一是找出在给定序列中缺失的两个数字，二是判断一个给定数字是否可以通过斐波那契数列的若干乘积来表示。通过详细解释解题思路并提供代码示例，本文旨在帮助读者理解和解决这类常见的编程问题。

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1001

Missing number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 748 Accepted Submission(s): 275

Problem Description

  
  
   
   There is a permutation without two numbers in it, and now you know what numbers the permutation has. Please find the two numbers it lose.

Input

  
  
   
   There is a number 
   
   
   
   
    
    
     
     T
    
     shows there are 
   
   
   
   
    
    
     
     T
    
     test cases below. (
   
   
   
   
    
    
     
     T≤10
    
    )
   
   
For each test case , the first line contains a integers 
   
   
   
   
    
    
     
     n
    
     , which means the number of numbers the permutation has. In following a line , there are 
   
   
   
   
    
    
     
     n
    
     distinct postive integers.(
   
   
   
   
    
    
     
     1≤n≤1,000
    
    )

Output

  
  
   
   For each case output two numbers , small number first.

Sample Input

Sample Output

题意：给你n个数，问哪两个数丢失。

解题思路：题意给的n个数是1~n+2之间的数，因此只需要将在1~n+2之间却不在给定的n个数的数找出来即可。

参考代码：

#include <iostream>
#include <string.h>
#include <iomanip>
#include <algorithm>
#include <cmath>
using namespace std;
typedef long long ll;
int main(){
    int n,t,a;
    bool used[1003];
    cin>>t;
    while (t--){
        cin>>n;
        memset(used,false,sizeof(used));
        for (int i=0;i<n;i++){
            cin>>a;
            used[a]=true;
        }
        int flag=0;
        for (int i=1;i<=n+2;i++){
            if (used[i]==false){
                flag++;
                if (flag==1)
                    cout<<i<<" ";
                if (flag==2)
                    cout<<i<<endl;
            }
        }
    }
    return 0;
}

1002

Fibonacci

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 996 Accepted Submission(s): 40

Problem Description

  
  
   
   Following is the recursive definition of Fibonacci sequence:
   
   

   
   
   
    $F i = ⎧ ⎩ ⎨ 01 F i - 1 + F i - 2 i = 0 i = 1 i > 1$ 
   
   
Now we need to check whether a number can be expressed as the product of numbers in the Fibonacci sequence.

Input

  
  
   
   There is a number 
   
   
   
   
    
    
     
     T
    
     shows there are 
   
   
   
   
    
    
     
     T
    
     test cases below. (
   
   
   
   
    
    
     
     T≤100,000
    
    )
   
   
For each test case , the first line contains a integers n , which means the number need to be checked. 
   
   

   
   
   
   
    
    
     
     0≤n≤1,000,000,000

Output

  
  
   
   For each case output "Yes" or "No".

Sample Input

Sample Output

  
  
   
   Yes
No
Yes

题意：给出fib数列，问任意给定的一个n是否可以是fib数列中的若干fib数的积；

解题思路：首先用一个数组将fib数列保存下来，然后求出用一个数组将Fibonacci数组中属于n的因子的数保存，最后在递归搜索求解是否存在n是这些Fibonacci数组成的积；

参考代码：

#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <stack>
#include <cmath>
#include <string.h>
using namespace std;
int fib[100],a[100],i,k;
bool work(int n,int step){	//递归搜索求解是否存在n是这些Fibonacci数组成的积
	if (n==1)
		return true;
	for (int j=step;j<k;j++){
		if (n%a[j]==0){
			if (work(n/a[j],j)==true)
				return true;
		}
	}
	return false;
}
int main(){
	int t,n;
	/*构造Fibonacci数组*/
	fib[0]=0;
	fib[1]=1;
	for (i=2;i<46;i++){
		fib[i]=fib[i-1]+fib[i-2];
	}
	cin>>t;
	while (t--){
		cin>>n;
		if (n==0){
			cout<<"Yes"<<endl;
			continue;
		}
		k=0;
		for (int j=3;j<46;j++){	//用一个数组将Fibonacci数组中属于n的因子的数保存
			if (n%fib[j]==0)
				a[k++]=fib[j];
		}
		if (work(n,0)==true)
			cout<<"Yes"<<endl;
		else
			cout<<"No"<<endl;
	}
	return 0;
}