2013-07-30 20:35 388人阅读 评论(0) 收藏 举报

http://acm.hdu.edu.cn/showproblem.php?pid=4627

The Unsolvable Problem

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 230    Accepted Submission(s): 136

Problem Description
There are many unsolvable problem in the world.It could be about one or about zero.But this time it is about bigger number.
Given an integer n(2 <= n <= 109).We should find a pair of positive integer a, b so that a + b = n and [a, b] is as large as possible. [a, b] denote the least common multiplier of a, b.
 

 

Input
The first line contains integer T(1<= T<= 10000),denote the number of the test cases.
For each test cases,the first line contains an integer n.
 

 

Output
For each test cases,print the maximum [a,b] in a line.


标准签到题，先特判n==2的情况，否则若n为奇数，则答案显然是n/2 和 n/2+1，若n为偶数设n=2*k，若k为偶数，则答案为k-1,k+1,否则答案为k-2,k+2。

[cpp] view plaincopyprint?
    1. #include <iostream>  
    2. #include <string.h>  
    3. #include <stdio.h>  
    4. #include <algorithm>  
    5. #define ll long long  
    6. using namespace std;  
    7. int main()  
    8. {  
    9.     //freopen("dd.txt","r",stdin);  
    10.     ll n,ncase;  
    11.     cin>>ncase;  
    12.     while(ncase--)  
    13.     {  
    14.         cin>>n;  
    15.         if(n==2)  
    16.         cout<<1<<endl;  
    17.         else  
    18.         {  
    19.             if(n%2)  
    20.             cout<<(n/2)*(n-n/2)<<endl;  
    21.             else  
    22.             {  
    23.                 if((n/2)%2==0)  
    24.                 cout<<(n/2-1)*(n/2+1)<<endl;  
    25.                 else  
    26.                 cout<<(n/2-2)*(n/2+2)<<endl;  
    27.             }  
    28.         }  
    29.     }  
    30.     return 0;  
    31. }
    32. 一开始用暴力，唉，挂了，但是为什么是这样呢，想了半天终于明白了。