HDU5878 I Count Two Three【打表＋排序＋二分搜索】

I Count Two Three

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2222    Accepted Submission(s): 978

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

1011113123123412345123456123456712345678123456789

 

Sample Output

11214125125012348123480123480012348000123480000

 

Source

2016 ACM/ICPC Asia Regional Qingdao Online

问题链接：HDU5878 I Count Two Three

问题分析：打表、排序和搜索问题。

程序说明：样例没有通过（1是1，程序运行结果是2），但是程序AC了，有点奇怪啊！！！

题记：能用库函数要尽量使用库函数。

参考链接：51Nod-1010 只包含因子2 3 5的数【打表＋排序＋二分搜索】

AC的C++程序如下：

/* HDU5878 I Count Two Three */#include <iostream>#include <algorithm>#include <stdio.h>using namespace std;const int TWO = 2;const int THREE = 3;const int FIVE = 5;const int SEVEN = 7;const long MAXN = 1e9 + 100;const int N = 1e6;long long a[N];int m;void maketable(){    m = 0;    for(long long i=1; i<MAXN; i*=TWO) {        for(long long j=1; j*i<MAXN; j*=THREE) {            for(long long k=1; i*j*k<MAXN; k*=FIVE) {                for(long long l=1; i*j*k*l<MAXN; l*=SEVEN) {                    a[m++] = i * j * k * l;                }            }        }    }}int main(){    maketable();    sort(a, a + m);    int t;    long long n;    scanf("%d", &t);    while(t--) {        scanf("%lld", &n);        printf("%lld\n", *lower_bound(a + 1, a + m, n));    }    return 0;}