UVA136-Ugly Numbers

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一道很经典的题，充分的利用了优先队列的各种性质，比起来用两个set，优先队列更清晰！

用set

#include<stdio.h>#include<set>#include<vector>#include<iostream>#include<queue>using namespace std;#define ll long longll a[3]={2,3,5};int main(){    set<ll>num;    set<ll>::iterator it;    set<ll>s;    num.insert(1);    s.insert(1);    for(int i=1;;i++){        it=num.begin();        ll x=*it;        num.erase(x);        if(i==1500){            cout<<"The 1500'th ugly number is "<<x<<".\n";            break;        }        for(int j=0;j<3;j++){            ll x2=x*a[j];            if(!s.count(x2)) {s.insert(x2);num.insert(x2); }        }    }}

一个set用来存放数字，一个用来反复的取和拿。

优先队列

#include<stdio.h>#include<set>#include<vector>#include<iostream>#include<queue>using namespace std;#define ll long longll a[3]={2,3,5};int main(){    priority_queue<ll,vector<ll>,greater<ll> > num;    set<ll>s;    num.push(1);    s.insert(1);    for(int i=1;;i++){        ll x=num.top();num.pop();        if(i==1500){            cout<<"The 1500'th ugly number is "<<x<<".\n";            break;        }        for(int j=0;j<3;j++){            ll x2=x*a[j];            if(!s.count(x2)) {s.insert(x2);num.push(x2); }        }    }}

简单明了的push，pop。

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