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ZOJ 2562 More Divisors

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2023-09-11 14:15:28 时间

More Divisors

Time Limit: 2000ms
Memory Limit: 65536KB
This problem will be judged on ZJU. Original ID: 2562
64-bit integer IO format: %lld      Java class name: Main

Everybody knows that we use decimal notation, i.e. the base of our notation is 10. Historians say that it is so because men have ten fingers. Maybe they are right. However, this is often not very convenient, ten has only four divisors -- 1, 2, 5 and 10. Thus, fractions like 1/3, 1/4 or 1/6 have inconvenient decimal representation. In this sense the notation with base 12, 24, or even 60 would be much more convenient.

The main reason for it is that the number of divisors of these numbers is much greater -- 6, 8 and 12 respectively. A good quiestion is: what is the number not exceeding n that has the greatest possible number of divisors? This is the question you have to answer.

Input:

The input consists of several test cases, each test case contains a integer n (1 <= n <= 1016).

Output:

For each test case, output positive integer number that does not exceed n and has the greatest possible number of divisors in a line. If there are several such numbers, output the smallest one.

Sample Input: 
10
20
100
Sample Output: 
6
12
60

 

Source

Andrew Stankevich's Contest #4

Author

Andrew Stankevich
 
解题：直接贪心地搜，选完2，一定从3开始选，不会直接从2跳到4，没必要，因为即使因子数相同，也会导致目标值变大
 
 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 typedef long long LL;
 4 const LL INF = 0x3f3f3f3f3f3f3f3f;
 5 const int maxn = 1010;
 6 int p[maxn],tot;
 7 bool np[maxn] = {true,true};
 8 void init(){
 9     tot = 0;
10     for(int i = 2; i < maxn; ++i){
11         if(!np[i]) p[tot++] = i;
12         for(int j = 0; j < tot && p[j]*i < maxn; ++j){
13             np[p[j]*i] = true;
14             if(i%p[j] == 0) break;
15         }
16     }
17 }
18 
19 LL ret,tsum,x;
20 
21 void dfs(int cur,LL sum,LL val){
22     if(sum > tsum || sum == tsum && val < ret){
23         ret = val;
24         tsum = sum;
25     }
26     for(int i = 1; i < 61; ++i){
27         val *= p[cur];
28         if(val > x) break;
29         dfs(cur + 1,sum*(i + 1),val);
30     }
31 }
32 int main(){
33     init();
34     while(~scanf("%lld",&x)){
35         ret = INF;
36         tsum = 1;
37         dfs(0,1,1);
38         printf("%lld\n",ret);
39     }
40     return 0;
41 }
View Code

 

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