HDU 5878

2016 ACM/ICPC Asia Regional Dalian Online 

I Count Two Three

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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 -th attacks failed if and only if  can be rewritten as the form of  which  are non-negative integers. 

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

Input

The first line of input contains an integer , the number of test cases.  test cases follow. Each test case provides one integer .

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

10
1
11
13
123
1234
12345
123456
1234567
12345678
123456789

Sample Output

1
12
14
125
1250
12348
123480
1234800
12348000
123480000

水题，t组测试样例，输入n，输出pow(2,a)*pow(3,b)*pow(5,c)*pow(7,d)的最小的大于等于n的数

先打表，再排序，最后用lower_bound函数找到最小的大于等于n的数的位置，然后输出即可

#include <iostream>
#include <cstdio>
#include <math.h>
#include <algorithm>
using namespace std;

int t;
long long n;
long long f[1000000];

int main()
{
    scanf("%d", &t);
    int cnt = 0;
    for(int a = 0; a <= 32; a++)
    {
        for(int b = 0; b <= 25; b++)
        {
            for(int c = 0; c <= 20; c++)
            {
                for(int d = 0; d <= 15; d++)
                {
                    if (a + b + c + d < 50) f[cnt++] = pow(2,a) * pow(3,b) * pow(5,c) * pow(7,d);
                }
            }
        }
    }
    sort(f, f + cnt);
    while(t--)
    {
        scanf("%lld", &n);
        long long tmp = lower_bound(f, f + cnt, n) - f;
        printf("%lld\n", f[tmp]);
    }
    return 0;
}