POJ2545-丑数

题目：题目链接

 

题意：就是按照丑数的定义，给定三个素数，然后以这三个数字为因子求地num个丑数：

 

丑数求法介绍：http://blog.youkuaiyun.com/acm_zl/article/details/10613073

 

#include <iostream>
#include <cstdio>
#include <string>
#include <string.h>
#include <map>
#include <vector>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <queue>
#include <set>
#include <stack>
#include <functional>
#include <fstream>
#include <sstream>
#include <iomanip>
#include <numeric>
#include <cassert>
#include <bitset>
#include <stack>
#include <ctime>
#include <list>
#define INF 0x7fffffff
#define max3(a,b,c) (max(a,b)>c?max(a,b):c)
#define min3(a,b,c) (min(a,b)<c?min(a,b):c)
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;

long long s[200000];
long long min(long long a,long long b,long long c)
{
    if(a > b)
    {
        if(b > c)
            return c;
        else
            return b;
    }
    else
    {
        if(a > c)
            return c;
        else
            return a;
    }
}
int main()
{
    int i,n,p1,p2,p3;
    long long t1,t2,t3;
    int i1,i2,i3;
    while(scanf("%d%d%d%d",&p1,&p2,&p3,&n)==4)
    {
        i1=1;
        i2=1;
        i3=1;
        s[1]=1;
        for(i = 2; i <= n+1; i++)
        {
            t1 = p1*s[i1];
            t2 = p2*s[i2];
            t3 = p3*s[i3];
            s[i]=min(t1,t2,t3);
            if(s[i]==t1)
                i1++;
            if(s[i]==t2)
                i2++;
            if(s[i]==t3)
                i3++;
        }
        printf("%lld\n",s[n+1]);
    }
    return 0;
}