1001 Maximum Multiple(2018 Multi-University Training Contest 1)

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Maximum Multiple

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3985    Accepted Submission(s): 926


Problem Description
Given an integer n, Chiaki would like to find three positive integers xy and z such that: n=x+y+zxnynzn and xyz is maximum.
 
Input
There are multiple test cases. The first line of input contains an integer T (1T106), indicating the number of test cases. For each test case:
The first line contains an integer n (1n106).
 
Output
For each test case, output an integer denoting the maximum xyz. If there no such integers, output ?1 instead.
 
Sample Input
3
1
2
3
 
Sample Output
-1
-1
1
 
题意:给定一个正整数n,满足n=x+y+z, 并且x , y , z能整除n,求满足条件的x,y,z的max(x,y,z)
 
分析: 首先理解整数1分成三个分子为1的分数和,仅有两种(1/3,1/3,1/3),(1/2,1/4,1/4)
所以要将n分解成三份并且能整除的话就只能分成(n/3,n/3,n/3),(n/2,n/4,n/4)
 
只需要判断n能不能被3或者4整除
 
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 int t;
 5 long long n;
 6 
 7 int main()
 8 {
 9     scanf("%d",&t);
10     while(t--) {
11         scanf("%d",&n);
12         if(n%3 == 0) {
13             printf("%lld
",n/3*n/3*n/3);
14         } else if(n%4 == 0) {
15             printf("%lld
",n/2*n/4*n/4);
16         } else {
17             printf("-1
");
18         }
19     }
20     
21     return 0;
22 }

 

 








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