hdu 5878 I Count Two Three【预处理打表+二分】

I Count Two Three

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 3582    Accepted Submission(s): 1094


[align=left]Problem Description[/align]
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.
 

[align=left]Input[/align]
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).
 

[align=left]Output[/align]
For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than
n.
 

[align=left]Sample Input[/align]
10

1

11

13

123

1234

12345

123456

1234567

12345678

123456789

 

[align=left]Sample Output[/align]
1

12

14

125

1250

12348

123480

1234800

12348000

123480000

题目大意：

给你一个数n，让你求一个大于等于n的最小的满足题意中2^a*3^b*5^c*7^d的数字。

思路：

1、很明显，2^32大于了1000000000，那么我们可以O（32^4）预处理出来所有满足这个条件的数。

2、那么我们再对每一个输入的数，二分查找最优解即可（大于等于n的最近数值）

Ac代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define ll __int64
ll ans[100000];
ll poww(ll di,ll ci)
{
if(ci==-1)return 0;
ll ans=1;
for(int i=1;i<=ci;i++)
{
ans*=di;
}
return ans;
}
void init()
{
int cont=0;
for(int i=0;i<32;i++)
{
if(poww(7,i)>1000000000)break;
for(int j=0;j<32;j++)
{
if(poww(7,i)*poww(5,j)>1000000000)break;
for(int k=0;k<32;k++)
{
if(poww(7,i)*poww(5,j)*poww(3,k)>1000000000)break;
for(int l=0;l<32;l++)
{
ll tmpp=poww(2,l)*poww(3,k)*poww(5,j)*poww(7,i);
if(tmpp>1000000000)break;
ans[cont++]=tmpp;
}
}
}
}
}
int main()
{
int t;
init();
sort(ans,ans+5999);
scanf("%d",&t);
while(t--)
{
ll n;
scanf("%I64d",&n);
int l=0,r=6000;
int mid;
ll anss;
while(r>=l)
{
int mid;
mid=(l+r)/2;
if(ans[mid]<n)
{
l=mid+1;
}
else
{
anss=ans[mid];
r=mid-1;
}
}
printf("%I64d\n",anss);
}
}