代码语言：javascript复制
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
代码语言：javascript复制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
代码语言：javascript复制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
代码语言：javascript复制10
1
11
13
123
1234
12345
123456
1234567
12345678
123456789Sample Output
代码语言：javascript复制1
12
14
125
1250
12348
123480
1234800
12348000
123480000这个题的意思是让你找到一个k大于n并且能写成 2 a 3 b 5 c 7 d 2^a3^b5^c7^d 2a3b5c7d这种形式。 然后就先把能写成这种形式的数的全部处理出来，在进行二分查找，比这个大的最小的数。
代码语言：javascript复制#include <bits/stdc .h>
using namespace std;
typedef long long ll;
long long a[200000];
int d = 0;
int main()
{
int x, y;
d = 0;
memset(a, 0, sizeof(a));
for (int i = 0; i <= 32; i )
{
for (int j = 0; j <= 19; j )
{
for (int k = 0; k <= 12; k )
{
for (int h = 0; h <= 11; h )
{
long long s = pow(2, i) * pow(3, j) * pow(5, k) * pow(7, h);
if (s > 1000000000 || s < 0)
break;
else
{
a[d ] = s;
}
}
}
}
}
sort(a, a d);
scanf("%d", &x);
while (x--)
{
scanf("%d", &y);
printf("%lldn", *lower_bound(a, a d, y));
}
return 0;
}
