2016 ACM/ICPC Asia Regional Qingdao Online 1001 I Count Two Three

Posted 2020-08-09 Annetree的博客

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem Description

I will show you the most popular board game in the Shanghai Ingress Resistance Team.
It all started several months ago.
We found out the home address of the enlightened agent Icount2three and decided to draw him out.
Millions of missiles were detonated, but some of them failed.

After the event, we analysed the laws of failed attacks.
It‘s interesting that the i-th attacks failed if and only if i can be rewritten as the form of 2a3b5c7d which a,b,c,d are non-negative integers.

At recent dinner parties, we call the integers with the form 2a3b5c7d "I Count Two Three Numbers".
A related board game with a given positive integer n from one agent, asks all participants the smallest "I Count Two Three Number" no smaller than n.

 

Input

The first line of input contains an integer t (1≤t≤500000), the number of test cases. t test cases follow. Each test case provides one integer n (1≤n≤109).

 

Output

For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than n.

 

Sample Input

10

1

11

13

123

1234

12345

123456

1234567

12345678

123456789

 

Sample Output

1

12

14

125

1250

12348

123480

1234800

12348000

123480000

 

先打表，然后二分搜索做

 1 #include <iostream>
 2 #include<cstdio>
 3 #include<algorithm>
 4 #include<cmath>
 5 #include<cstdlib>
 6 #include<set>
 7 using namespace std;
 8 int t,x,len;
 9 int arr[6000];
10 int main()
11 {
12     len=0;
13   for(int a=0;a<=32;a++)
14   {
15       double aa=pow(2.0,a*1.0);
16       for(int b=0;b<=19;b++)
17       {
18           double bb=pow(3.0,b*1.0);
19           if (aa*bb>=1200000000.0) break;
20           for(int c=0;c<=14;c++)
21           {
22               double cc=pow(5.0,c*1.0);
23               if (aa*bb*cc>=1200000000.0) break;
24               for(int d=0;d<=11;d++)
25               {
26                   double dd=pow(7.0,d*1.0);
27                   if (aa*bb*cc*dd>=1200000000.0) break;
28                   arr[++len]=(int)aa*bb*cc*dd;
29               }
30           }
31       }
32   }
33   sort(arr+1,arr+len+1);
34   //printf("%d\n",arr[len]);
35     //for(int i=1;i<=len;i++) printf("%d ",arr[i]);
36    // printf("%d\n",len);
37     while(scanf("%d",&t)!=EOF)
38     {
39         for(;t>0;t--)
40         {
41             scanf("%d",&x);
42 
43 //            set<int> s;
44 //            s.insert(1);
45 //            while(!s.empty())
46 //            {
47 //                set<int>::iterator ii;
48 //                ii=s.begin();
49 //                if (*ii>=x) {printf("%d\n",*ii); break;}
50 //                s.erase(s.begin());
51 //                int k=*ii;
52 //                if(k*2<1200000000) s.insert(k*2);
53 //                if(k*3<1200000000) s.insert(k*3);
54 //                if(k*5<1200000000) s.insert(k*5);
55 //                if(k*7<1200000000) s.insert(k*7);
56 //            }
57 
58             int l=1,r=len;
59             while(l<r)
60             {
61               int mid=(l+r)/2;
62               if (arr[mid]<x) l=mid+1;
63                 else r=mid;
64             }
65             printf("%d\n",arr[l]);
66 
67         }
68     }
69     return 0;
70 }