hdu 5878

I Count Two Three

Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2719 Accepted Submission(s): 1177

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

Input

The first line of input contains an integer

Output

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

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

Source

2016 ACM/ICPC Asia Regional Qingdao Online

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <algorithm>
 4 #include <cstdlib>
 5 #include <cstring>
 6 #include <string>
 7 #include <deque>
 8 #include <vector>
 9 #include <set>
10 using namespace std;
11 #define ll long long 
12 #define P pair<int,int>
13 const int N= 5e3+1000;
14 const ll  M = 1000000000;
15 ll n,a[N];
16 int cnt =0;
17 int t;
18 ll pow(ll a,ll b){
19     ll ans=1;
20     while(b){
21         if(b&1) ans=ans*a;
22         b>>=1;
23         a*=a;
24     }
25     return ans;
26 }
27 void init()
28 {
29     ll i,j,k,l;
30     for(i=0;i<=32;i++){
31         if(pow(2,i)>M) break;
32         for(j=0;j<=32;j++){
33             if(pow(2,i)*pow(3,j)>M) break;
34             for(k=0;k<=32;k++) {//k++后，l 再次从0开始。 
35                 if(pow(2,i)*pow(3,j)*pow(5,k)>M) break;
36                 for(l=0;l<=32;l++){
37                     if(pow(2,i)*pow(3,j)*pow(5,k)*pow(7,l)>M) break;//在这里跳出才可以继续循环 
38                     a[cnt++] = pow(2,i)*pow(3,j)*pow(5,k)*pow(7,l);
39                 }
40             }
41         }
42     }
43     sort(a,a+cnt);
44 }
45 int main()
46 {
47     
48     init();
49     scanf("%d",&t);
50     while(t--){
51         scanf("%lld",&n);
52         printf("%lld\n",a[lower_bound(a,a+cnt,n)-a]);
53     }
54     return 0;
55 }