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

2
3
3 4 5
1
1

Sample Output

1 2
2 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

3
4
17
233

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;
}