hdu 5878 I Count Two Three

I Count Two Three

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 719    Accepted Submission(s): 380

Problem DesI will show you the most popular board game in the Shanghai Ingress Resistance TeIt all started several months agoWe 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 failedAfter 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 2^a*3^b*5^c*7^dwhich a,b,c,d are non-negative integers.

At recent dinner parties, we call the integers with the form2^a*3^b*5^c*7^d "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≤10^9) .

 

 

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

 

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online 

 

题意：对于每一个输入的N，找一个不小于N的最小的满足条件的值（使得这个值的因子是2,3,5,7）

思路：暴力，之后二分搜索来查找。

AC代码：

#define _CRT_SECURE_NO_DEPRECATE
#include<iostream>
#include<algorithm>
using namespace std;
const int maxn = 1000000000 + 5;
const int N_MAX =5200;
typedef long long LL;
//LL table[10000];
LL a[N_MAX] = {0,1,2,3,4,5,6,7,8,9,10};
LL p[4] = { 2,3,5,7 };
int main() {
    int T,n;
    for (int i = 10;i < N_MAX;i++) {
        a[i] = INT_MAX;
        int k = i - 1;
        for (int j = 0;j < 4;j++) {
            while (a[k-1] * p[j]>a[i - 1])
                k--;
            a[i] = min(a[i],a[k]*p[j]);
        }
    }
    /*int cnt = 0;
    for (LL a =1;a <maxn;a*=2) {
        for (LL b=1;a*b <maxn;b*=3) {
            for (LL c = 1;a*b*c < maxn;c *= 5) {
                for (LL d = 1;a*b*c*d < maxn;d *= 7) {
                    table[cnt++] = a*b*c*d;
                }
            }
        }
    }
    sort(table, table + cnt);*/
    scanf("%d",&T);
    while (T--) {
        scanf("%d",&n);
        int id=*lower_bound(a+1, a +N_MAX,n);
        printf("%d\n", id);
    }
    
    return 0;
}