Maximum Multiple Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3241 Accepted Submission(s): 134 ...
Maximum Multiple
Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3241 Accepted Submission(s): 1344
Input There are multiple test cases. The first line of input contains an integer T (1≤T≤106), indicating the number of test cases. For each test case:
The first line contains an integer n (1≤n≤106).
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 題目大意：n = x+ y + z, 需要滿足 x|n, y|n, z|n (整除關係)，輸出xyz的最大值，若不存在，輸出-1 設 n/x = r n/y = s n/z = t ， r <= s <= t 所以1/r + 1/s + 1/t = 1 所以r <= 3 當r=3 s,t = (3,3) 當r=2 s,t = (4, 4) , (3, 6) 所以三種情況 n/3 n/3 n/3 n/2 n/4 n/4 n/2 n/3 n/6 (沒有第一種大，捨去) 所以就判是否能 整除3 或 4
1 #include <iostream> 2 #include <stdio.h> 3 #include <math.h> 4 #include <string.h> 5 #include <stdlib.h> 6 #include <string> 7 #include <vector> 8 #include <set> 9 #include <map> 10 #include <queue> 11 #include <algorithm> 12 #include <sstream> 13 #include <stack> 14 using namespace std; 15 #define rep(i,a,n) for (int i=a;i<n;i++) 16 #define per(i,a,n) for (int i=n-1;i>=a;i--) 17 #define pb push_back 18 #define mp make_pair 19 #define all(x) (x).begin(),(x).end() 20 #define fi first 21 #define se second 22 #define SZ(x) ((int)(x).size()) 23 #define FO freopen("in.txt", "r", stdin) 24 #define lowbit(x) (x&-x) 25 #define mem(a,b) memset(a, b, sizeof(a)) 26 typedef vector<int> VI; 27 typedef long long ll; 28 typedef pair<int,int> PII; 29 const ll mod=1000000007; 30 const int inf = 0x3f3f3f3f; 31 ll powmod(ll a,ll b) {ll res=1;a%=mod;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;} 32 ll gcd(ll a,ll b) { return b?gcd(b,a%b):a;} 33 //head 34 int _, n; 35 int main() { 36 for(scanf("%d", &_);_;_--) { 37 scanf("%d", &n); 38 if(n%3 == 0) printf("%lld\n", 1ll * n * n * n / 27); 39 else if(n%4 == 0) printf("%lld\n", 1ll * n * n * n / 32); 40 else puts("-1"); 41 } 42 }
