18005 It is not ugly number

18005 It is not ugly number

时间限制:2000MS 内存限制:65535K

提交次数:0 通过次数:0

题型: 编程题 语言: G++;GCC

Description

Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence1, 2, 3, 4, 5, 6, 8, 9, 10, 12, …shows the

first 10 ugly numbers. By convention, 1 is included. Then, here are the first 10 Not ugly numbers:7, 11, 13, 14, 17, 19,

21, 22, 23, 26. Given the integer n, write a program to find and print the n’th Not ugly number.

输入格式

First line is T（T<=10000）, the number of cases.

The T lines following. Each line of the input contains a positive integer n (n <= 100000000).

输出格式

For each case, output the n’th Not ugly number .

输入样例

3

1

2

9

输出样例

7

11

23

#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
int min3(int a,int b,int c)
{
int min=a>b?b:a;
return min>c?c:min;
}
int main()
{
int min,*p,i,j,f2=0,f3=0,f5=0,n=1500,count;
p=(int *)malloc(n*sizeof(int ));
*p=1;

for(i=1;i<n;i++)
{
min=min3(*(p+f2)*2,*(p+f3)*3,*(p+f5)*5);
p[i]=min;
if(min==*(p+f2)*2) f2++;
if(min==*(p+f3)*3) f3++;
if(min==*(p+f5)*5) f5++;
}
cin>>j;
while(j--)
{
scanf("%d",&n);
i=-1,count=0;
while(i++,1)
{
count+=p[i+1]-p[i]-1;
if(count>=n) break;
}
printf("%d\n",p[i+1]+n-count-1);
}
}