2016 ACM/ICPC Asia Regional Qingdao Online 1001/HDU5878 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): 782 Accepted Submission(s): 406
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.
/******************************
code by drizzle
blog: www.cnblogs.com/hsd-/
^ ^ ^ ^
O O
******************************/
#include<bits/stdc++.h>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cstdio>
#define ll long long
#define mod 1000000007
#define PI acos(-1.0)
#define N 1000000000
using namespace std;
int t;
ll a2[]={},a3[]={},a5[]={},a7[]={};
int jishu;
ll ans[];
void init()
{
jishu=;
int er=,san=,wu=,qi=;
for(int i=; a2[i-]<=N;er++,i++)
a2[i]=a2[i-]*;
for(int i=; a3[i-]<=N;san++,i++)
a3[i]=a3[i-]*;
for(int i=; a5[i-]<=N;wu++,i++)
a5[i]=a5[i-]*;
for(int i=; a7[i-]<=N;qi++, i++)
a7[i]=a7[i-]*;
for(int i=; i<er; i++)
for(int j=; a2[i]*a3[j]<=N&&j<san; j++)
for(int k=; a2[i]*a3[j]*a5[k]<=N&&k<wu; k++)
for(int l=; a2[i]*a3[j]*a5[k]*a7[l]<=N&&l<qi; l++)
ans[jishu++]=a2[i]*a3[j]*a5[k]*a7[l];
sort(ans,ans+jishu);
}
int main()
{
int n;
while(scanf("%d",&t)!=EOF)
{
init();
for(int i=; i<=t; i++)
{
scanf("%d",&n);
printf("%I64d\n",ans[lower_bound(ans,ans+jishu,n)-ans]);
}
}
return ;
}
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