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.

This problem statement is the exclusive and proprietary property of TopCoder, Inc. Any unauthorized use or reproduction of this information without the prior written consent of TopCoder, Inc. is strictly prohibited. (c)2003, TopCoder, Inc. All rights reserved.     

【題解】

分解質因數後判斷究竟是否相同。

只需以爲枚舉。

【代碼】

#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);
}