topcoder SRM 500 div2 level3

Problem Statement
	NOTE: This problem statement contains superscripts that may not display properly if viewed outside of the applet. You are given two geometric progressions S1 = (b1*q1i, 0 <= i <= n1-1) and S2 = (b2*q2i, 0 <= i <= n2-1). Return the number of distinct integers that belong to at least one of these progressions.
Definition
	Class: GeometricProgressions Method: count Parameters: int, int, int, int, int, int Returns: int Method signature: int count(int b1, int q1, int n1, int b2, int q2, int n2) (be sure your method is public)
Constraints
-	b1 and b2 will each be between 0 and 500,000,000, inclusive.
-	q1 and q2 will each be between 0 and 500,000,000, inclusive.
-	n1 and n2 will each be between 1 and 100,500, inclusive.
Examples
0)
	3 2 5 6 2 5 Returns: 6 The progressions in this case are (3, 6, 12, 24, 48) and (6, 12, 24, 48, 96). There are 6 integers that belong to at least one of them: 3, 6, 12, 24, 48 and 96.
1)
	3 2 5 2 3 5 Returns: 9 This time the progressions are (3, 6, 12, 24, 48) and (2, 6, 18, 54, 162). Each of them contains 5 elements, but integer 6 belongs to both progressions, so the answer is 5 + 5 - 1 = 9.
2)
	1 1 1 0 0 1 Returns: 2 The progressions are (1) and (0).
3)
	3 4 100500 48 1024 1000 Returns: 100500 Each element of the second progression belongs to the first progression as well.

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【题解】

分解质因数后判断究竟是否相同。

只需以为枚举。

【代码】

#include <vector>
#include <list>
#include <map>
#include <set>
#include <deque>
#include <stack>
#include <bitset>
#include <algorithm>
#include <functional>
#include <numeric>
#include <utility>
#include <sstream>
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <ctime>
#include <cstring>

using namespace std;

int a[100][3],b[100][3],t[100][3];
class GeometricProgressions
{
public:
	int count(int, int, int, int, int, int);
};

void prime(int a[][3],int &tot,int x,int y)
{
    tot=0;
    int i;
    for (i=2;i*i<=max(x,y);i++)
    {
        if (x<=1 && y<=1) return;
        if (x && x%i==0)
        {
            a[++tot][0]=i;
            while (x%i==0)
            {
                x/=i;
                a[tot][1]++;
            }
        }
        if (y && y%i==0)
        {
            if (a[tot][0]!=i) a[++tot][0]=i;
            while (y%i==0)
            {
                y/=i;
                a[tot][2]++;
            }
        }
    }
    if (x<y)
    {
        if (x!=1)
            a[++tot][0]=x,a[tot][1]=1;
        if (y!=1)
        {
            if (a[tot][0]!=y) a[++tot][0]=y;
            a[tot][2]=1;
        }
    }
    else {

    if (y!=1)
        a[++tot][0]=y,a[tot][2]=1;
    if (x!=1)
    {
        if (a[tot][0]!=x) a[++tot][0]=x;
        a[tot][1]=1;
    }

}
}

int GeometricProgressions::count(int b1, int q1, int n1, int b2, int q2, int n2)
{
    int i,j,t1,t2,ans,tt;
    bool ff;
    if (b2==0 || q2<=1)
    {
        swap(b1,b2);swap(q1,q2);swap(n1,n2);
    }
    if (b1==0 || q1<=1)
    {
        set<long long> v;
        v.insert(b1);
        if (n1>0) v.insert(b1*q1);
        long long cur=q2;
        for (i=0;i<n2;i++)
        {
            v.insert(b2*cur);
            cur*=q2;
            if (cur>500000000)
                return v.size()+n2-i-1;
        }
        return v.size();
    }
    prime(a,t1,q1,b1);
    prime(b,t2,q2,b2);
    ans=n1+n2;
    if (t1!=t2) return ans;
    for (i=1;i<=t1;i++)
        if (a[i][0]!=b[i][0]) return ans;
    if (b[1][1]==0)
    {
        swap(n1,n2);
        memcpy(t,a,sizeof(a));
        memcpy(a,b,sizeof(b));
        memcpy(b,t,sizeof(t));
    }
    for (i=0;i<n1;i++)
    {
        tt=a[1][1]*i+a[1][2]-b[1][2];
        if (tt%b[1][1]!=0) continue;
        if (tt<0) continue;
        tt/=b[1][1];
        if (tt>=n2) continue;
        ff=true;
        for (j=2;j<=t1;j++)
            if (a[j][1]*i+a[j][2]!=b[j][1]*tt+b[j][2])
            {
                ff=false;
                break;
            }
        if (ff) ans--;
    }
    return ans;
}
int main()
{
    GeometricProgressions a;
    cout << a.count(3,4,100500,48,1024,1000);
}