POJ 3993 Not So Flat After All

原来就是对两个数分别因数分解，然后分别放到两个集合里，求合并后的集合的大小。再求每个数与另一个集合中该数的幂级数之差，如果另一个集合中没有这个数，就相当于这个集合中这个数的幂级数是0，最后把这些差的绝对值相加就是结果。

/*
ID: sdj22251
PROG: calfflac
LANG: C++
*/
#include <iostream>
#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <queue>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <cctype>
#include <string>
#include <cstring>
#include <cmath>
#include <ctime>
#define MAX 100000000
#define LOCA
#define PI acos(-1.0)
using namespace std;
bool tag[1000011];
int p[1000011];
int cnt;
void get_prime()
{
    cnt = 0;
    tag[1] = 1;
    for (int i = 2; i < 1000005; i++)
    {
        if (!tag[i])
        p[cnt++] = i;
        for (int j = 0; j < cnt && p[j] * i < 1000005; j++)
        {
            tag[i*p[j]] = 1;
            if (i % p[j] == 0)
            break;
        }
    }
}
int main()
{
#ifdef LOCAL
    freopen("ride.in","r",stdin);
    freopen("ride.out","w",stdout);
#endif
    int a, b, i, cas = 0;
    get_prime();
    while(scanf("%d%d", &a, &b) != EOF)
    {
        if(a == 0 && b == 0) break;
        int ans = 0;
        int count = 0;
        for(i = 0; i < cnt && (p[i] <= a || p[i] <= b); i++)
        {
            bool fg1 = false;
            bool fg2 = false;
            int ct1 = 0;
            int ct2 = 0;
            if(a % p[i] == 0)
            {
                fg1 = true;
                while(a % p[i] == 0)
                {
                    a /= p[i];
                    ct1++;
                }
            }
            if(b % p[i] == 0)
            {
                fg2 = true;
                while(b % p[i] == 0)
                {
                    b /= p[i];
                    ct2++;
                }
            }
            if(fg1 || fg2)
            {
                count++;
                ans += abs(ct1 - ct2);
            }
        }
        printf("%d. %d:%d\n", ++cas, count, ans);
    }
    return 0;
}