hdu 4627 The Unsolvable Problem（暴力的搜索）

Posted 2020-06-08

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.

 

 

Sample Input

3 2 3 4

 

 

Sample Output

1 2 3

 

 

Author

WJMZBMR

 

 

Source

2013 Multi-University Training Contest 3

 

题意：给定一个数n，要求是找到一对数a、b，使得a+b=n，且a和b的最小公倍数要最大，求最大的最小公倍数。

思路：直接暴力查询，找出最大的值。具体看代码。时间复杂度是O（n），并不会超时。注意要用long long，会爆int。

 1 #pragma comment(linker, "/STACK:1024000000,1024000000")
 2 #include<iostream>
 3 #include<cstdio>
 4 #include<cstring>
 5 #include<cmath>
 6 #include<math.h>
 7 #include<algorithm>
 8 #include<queue>
 9 #include<set>
10 #include<bitset>
11 #include<map>
12 #include<vector>
13 #include<stdlib.h>
14 using namespace std;
15 #define ll long long
16 #define eps 1e-10
17 #define MOD 1000000007
18 #define N 1000000
19 #define inf 1e12
20 ll n;
21 ll gcd(ll a,ll b){
22     return b==0?a:gcd(b,a%b);
23 }
24 ll lcm(ll a,ll b){
25     return a*b/gcd(a,b);
26 }
27 int main()
28 {
29     ll t;
30     scanf("%I64d",&t);
31     while(t--){
32         scanf("%I64d",&n);
33         
34         ll x=n/2;
35         ll r=gcd(x,n-x);
36         ll ans=x*(n-x)/r;
37         while(r!=1 && x>0){
38             x--;
39             r=gcd(x,n-x);
40             ans=max(ans,x*(n-x)/r);
41         }
42         printf("%I64d\n",ans);
43     }
44     return 0;
45 }

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