# I Count Two Three HDU – 5878 (打表+二分搜索+输入挂)

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 ii-th attacks failed if and only if ii can be rewritten as the form of 2a3b5c7d2a3b5c7d which a,b,c,da,b,c,d are non-negative integers.

At recent dinner parties, we call the integers with the form 2a3b5c7d2a3b5c7d “I Count Two Three Numbers”.
A related board game with a given positive integer nn from one agent, asks all participants the smallest “I Count Two Three Number” no smaller than nn.

InputThe first line of input contains an integer t (1t500000)t (1≤t≤500000), the number of test cases. tt test cases follow. Each test case provides one integer n (1n109)n (1≤n≤109).OutputFor each test case, output one line with only one integer corresponding to the shortest “I Count Two Three Number” no smaller than nn.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

#include<iostream>
#include<algorithm>
using namespace std;
inline bool scan_d(int &num)
{
char in;bool IsN=false;
in=getchar();
if(in==EOF) return false;
while(in!='-'&&(in<'0'||in>'9')) in=getchar();
if(in=='-'){ IsN=true;num=0;}
else num=in-'0';
while(in=getchar(),in>='0'&&in<='9'){
num*=10,num+=in-'0';
}
if(IsN) num=-num;
return true;
}
long long sum(int a,int b,int c,int d){
long long x=1;
for(int i=0;i<a;i++){
x*=2;
}
for(int i=0;i<b;i++){
x*=3;
}
for(int i=0;i<c;i++){
x*=5;
}
for(int i=0;i<d;i++){
x*=7;
}
return x;

}
long long ax[8000] ;
int main(){
int s=0;
for(int a=0;a<30;a++){
for(int b=0;b<20;b++){
for(int c=0;c<15;c++){
for(int d=0;d<13;d++){
long long x=sum(a,b,c,d);
if(x<10000000000&&x>0){
ax[s++]=x;
}else{
break;
}
}
}
}
}

sort(ax,ax+s);
//	cout<<s<<endl;
int t;
scan_d(t);
int xxx;
for(int i=0;i<t;i++){
scan_d(xxx);
int L=0,R=s;
while(L<R){
int mid=(L+R)/2;
if(ax[mid]>=xxx){
R=mid;
}else{
L=mid+1;
}
}

printf("%d\n",ax[L]);
}

return 0;
}