Ugly Numbers(1.5.8)
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Ugly Numbers(1.5.8)
Time Limit:1000MS Memory Limit:10000KB 64bit IO Format:%I64d & %I64
Description
Ugly numbers are numbers whose only prime factors are 2, 3 or 5. The sequence
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, ...
shows the first 10 ugly numbers. By convention, 1 is included.
Given the integer n,write a program to find and print the n'th ugly number.
1, 2, 3, 4, 5, 6, 8, 9, 10, 12, ...
shows the first 10 ugly numbers. By convention, 1 is included.
Given the integer n,write a program to find and print the n'th ugly number.
Input
Each line of the input contains a postisive integer n (n <= 1500).Input is terminated by a line with n=0.
Output
For each line, output the n’th ugly number .:Don’t deal with the line with n=0.
Sample Input
1290
Sample Output
1210
题意:
把质因子只有2,3,5的数称为Ugly数,求第k大的Ugly数。(1是第一个)
思路:
定义优先队列,数越小优先级越高,先出队列,然后*2 *3 *5进队列(注意消去重复的数)
#include<stdio.h>#include<iostream>#include<queue>using namespace std;#define ll __int64struct node {ll num;friend bool operator < (const node &a,const node &b)//定义优先级{return a.num>b.num;}};ll bfs(ll n) {ll sum;node x,y,z;sum=0;priority_queue<node>que;x.num=1;z.num=0; que.push(x);// 1进队列while(1){x=que.top();que.pop();//优先级最高的先出队列if(x.num==z.num) continue;//与之前重复的数不计sum++;if(sum==n) return x.num;//找到第n个y.num=x.num*2;que.push(y);y.num=x.num*3;que.push(y);y.num=x.num*5;que.push(y);z.num=x.num;}return 0;}int main(){ll num;while(scanf("%I64d",&num)&&num){printf("%I64d\n",bfs(num));}return 0;}
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