HDU 5878-I Count Two Three（可被2 3 5 7整除的数）

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

Problem Description

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 thann.

 

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≤109).

 

Output

For each test case, output one line with only one integer corresponding to the shortest "I Count Two Three Number" no smaller than .

 

Sample Input

1011113123123412345123456123456712345678123456789

 

Sample Output

11214125125012348123480123480012348000123480000

 

题目意思：

求出一个最小的不小于N的可被2 3 5 7整除的数。

解题思路：

打表二分搜索，注意表的范围不能开得太大也不能开得太小。

#include<cstdio>#include<iostream>#include<vector>#include<cmath>#include<cstring>#include<queue>#include<algorithm>#include<set>#include<queue>#define N 10000using namespace std;const int maxn=200000*5;typedef long long ll;#define INF 0xfffffffll u[N];ll getmin(ll a,ll b){    return a<b?a:b;}void judge(int n){    u[0]=1;    int i2=0,i3=0,i5=0,i7=0;    int i=1;    while(i<n)    {        ll value=getmin(u[i2]*2,getmin(u[i3]*3,getmin(u[i5]*5,u[i7]*7)));        if(value==u[i2]*2)            ++i2;        if(value==u[i3]*3)            ++i3;        if(value==u[i5]*5)            ++i5;        if(value==u[i7]*7)            ++i7;        u[i++]=value;    }    //return u[n-1];}int my_lower_bound(ll num){    int l=0;    int h=N-1;    while(l<=h)    {        int mid=(l+h)/2;        if(num>=u[mid])            l=mid+1;        else            h=mid-1;    }    return l;}int main(){    judge(N);    int t;    ll n;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        int tmp=lower_bound(u,u+N-1,n)-u;        printf("%I64d\n",u[tmp]);    }    return 0;}