Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 537 Accepted Submission(s): 271
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 thei -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 form2a3b5c7d "I
Count Two Three Numbers".
A related board game with a given positive integern from
one agent, asks all participants the smallest "I Count Two Three Number" no smaller than n .
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
At recent dinner parties, we call the integers with the form
A related board game with a given positive integer
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
Source
【题解】
10^9用2乘就是2^30左右。用3 是3^19,又是5^13,又是7^11;
然后30*19*13*11不会很大！
注意2^0,3^0。。。
然后枚举2,3,5,7的指数。得到一个数字。
我是觉得不会有重复的数字。
但是有也没关系。lower_bound的功能足够解决这道题。
【代码】
#include <cstdio>
#include <algorithm>
const long long num[5] = { 0,2,3,5,7 };
const int limit[5] = { 0,30,19,13,11 };
const long long max = 1000000000;
long long a[5] = { 0 };
long long pre[5][31];
long long tt = 1;
long long b[90000];
int t, n = 0;
//max==9
void dfs(int now)//dfs枚举质因子的指数
{
if (tt > max)
return;
if (now == 5)
{
n++;
b[n] = tt;
return;
}
for (int i = 0; i <= limit[now]; i++)
{
long long kk = tt;
tt *= pre[now][i];
dfs(now + 1);
tt = kk;
}
}
int main()
{
//freopen("F:\rush.txt", "r", stdin);
//freopen("F:\rush_out.txt", "w", stdout);
for (int i = 1; i <= 4; i++)
pre[i][0] = 1;
for (int i = 1; i <= 4; i++)
{
for (int j = 1; j <= limit[i]; j++)//预处理出2^x,3^y...
pre[i][j] = pre[i][j - 1] * num[i];
}
dfs(1);
std::sort(b + 1, b + 1 + n);//排序。之后才能用lower_bound;
scanf("%d", &t);
long long query;
for (int i = 1; i <= t; i++)
{
scanf("%I64d", &query);
printf("%I64d
", b[std::lower_bound(b + 1, b + 1 + n, query) - b]);
}
return 0;
}