hdu5878-I Count Two Three

最新推荐文章于 2018-08-30 13:36:13 发布

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最新推荐文章于 2018-08-30 13:36:13 发布

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原文链接：http://www.cnblogs.com/wangtao971115/p/10358315.html

本文介绍了一个源自Ingress社区的数学问题，该问题要求找出最小的“ICountTwoThreeNumber”（形式为2^a*3^b*5^c*7^d的整数），且这个数不小于给定的正整数n。文章提供了使用预计算和二分搜索相结合的方法来高效求解这一问题的C++代码实现。

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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 than n .

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 n .

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

这道题目由于是2,3,5,7的次方,所以可能性不会太多,可以通过打表,然后二分查找的方法来解决

AC:

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
long long ans[1024000]={0};

int main()
{
    int t,cnt=1;
    scanf("%d",&t);
    long long n;
    for(int i=0;i<=30;i++)
            for(int j=0;j<=20;j++)
                for(int k=0;k<=15;k++)
                    for(int l=0;l<=11;l++)
                        if(i+j+k+l<32){
                            ans[cnt++]=pow(2,i)*pow(3,j)*pow(5,k)*pow(7,l);
                        }//打表后排序
    sort(ans,ans+cnt);
    while(t--)
    {
        scanf("%lld",&n);
        long long tmp=lower_bound(ans,ans+cnt,n)-ans;//二分查找大于或等于n的第一个数的位置
        printf("%lld\n",ans[tmp]);
    }
    return 0;
}

转载于:https://www.cnblogs.com/wangtao971115/p/10358315.html