BestCoder Round #28

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

233 4 511

 

Sample Output

1 22 3

 

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:
Fi=⎧⎩⎨01Fi−1+Fi−2i = 0i = 1i > 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

3417233

 

Sample Output

YesNoYes

 

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;elsecout<<"No"<<endl;}return 0;}