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>
#include <cmath>
#include <string.h>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>

using namespace std;

int a[1500];
typedef pair<int,int> temp;
int main()
{
    int t,n,i,cnt=0;
    priority_queue<temp,vector<temp>,greater<temp> > pq;
    temp tempp;
    pq.push(temp(1,2));
    for(i=0;i<1205;++i){
        tempp=pq.top();
        switch(tempp.second){
            case 2:pq.push(temp(tempp.first*2,2));
            case 3:pq.push(temp(tempp.first*3,3));
            case 5:pq.push(temp(tempp.first*5,5));
        }
        a[i]=tempp.first;
        pq.pop();
    }
    scanf("%d",&t);
    while(t--){
        scanf("%d",&n);
        cnt=0;
        i=0;
        while(cnt<n){
            cnt+=a[i+1]-a[i++]-1;
        }
        i--;
        cnt-=a[i+1]-a[i]-1;
        printf("%d\n",a[i]+n-cnt);
    }
    return 0;
}