hdu 1695 GCD （歐拉函數、容斥原理）

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 7357    Accepted Submission(s): 2698

Problem Description Given 5 integers: a, b, c, d, k, you're to find x in a...b, y in c...d that GCD(x, y) = k. GCD(x, y) means the greatest common divisor of x and y. Since the number of choices may be very large, you're only required to output the total number of different number pairs.

Please notice that, (x=5, y=7) and (x=7, y=5) are considered to be the same.

Yoiu can assume that a = c = 1 in all test cases.

Input The input consists of several test cases. The first line of the input is the number of the cases. There are no more than 3,000 cases.

Each case contains five integers: a, b, c, d, k, 0 < a <= b <= 100,000, 0 < c <= d <= 100,000, 0 <= k <= 100,000, as described above.

Output For each test case, print the number of choices. Use the format in the example.

Sample Input

2
1 3 1 5 1
1 11014 1 14409 9
        

Sample Output

Case 1: 9
Case 2: 736427


   
    
     Hint
    For the first sample input, all the 9 pairs of numbers are (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (2, 3), (2, 5), (3, 4), (3, 5).
   
    
        

Source 2008 “Sunline Cup” National Invitational Contest

題目大意：求出[a,b]和[c,d]區間裡面gcd（x,y）=k的數的對數。 思路：既然是求gcd為k的數的對數，不妨先将b和d都除以k，這樣問題就轉化為[1,n]和[1,m]區間裡面gcd（x,y）為1 的數的對數。因為題目裡已經說明a和c 可以認為是1，這樣就更簡單了。對于一個[1,n]的區間，我們可以用歐拉函數算出總對數。 那麼問題就可以分解成2個： 1、在[1,n]上用歐拉函數算出總對數。 2、在[n+1,m]上，計算在[1,n]裡面的總對數，可以用容斥原理。

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<iostream>
#include<algorithm>
#include<vector>
#define min(a,b) a<b?a:b
#define max(a,b)  a>b?a:b
#define Max 100005
#define LL __int64
using namespace std;
LL sum[Max],tot;
int p[Max][20];
int num[Max];
void init()
{
	sum[1]=1;
	for(int i=2;i<Max;i++)
	sum[i]=i;
	for(int i=2;i<Max;i++)
	if(sum[i]==i)
	for(int j=i;j<Max;j+=i)
	sum[j]=sum[j]/i*(i-1);
	
}
void init2()
{
	 LL x,k,i,j;
	for( i=1;i<=Max;i++)
	{
		x=i;k=0;
		for(j=2;j<=sqrt(i);j++)
		{
			if(x%j==0){
				while(x%j==0)x=x/j;
			//	p[i].push_back(j);
				p[i][num[i]++]=j;
			}
		}
		if(x>1)p[i][num[i]++]=x;
	}
}
LL dfs(int n,int b,int x,int k)
{
	LL ans=0;
	for(int i=x;i<k;i++)
	{
		ans+=b/p[n][i]-dfs(n,b/p[n][i],i+1,k);
	}
	return ans;
}
int main()
{
	LL T,a,b,c,d,k;
	int i,j,t;
	init();
	init2();
//	printf("%I64d %I64d\n",sum[2],sum[3]);
	scanf("%I64d",&T);
	t=0;
	while(T--)
	{
		tot=0;
		t++;
		scanf("%I64d%I64d%I64d%I64d%I64d",&a,&b,&c,&d,&k);
		printf("Case %d: ",t);
		if(k==0){printf("0\n");continue;}
		b=b/k;
		d=d/k;
		int m;
		m=min(b,d);
		d=max(b,d);
		b=m;
		for(i=1;i<=b;i++)
		tot=tot+sum[i];
	   for(i=b+1;i<=d;i++)
	   {
	 //  printf("%d\n",p[i].size());
	   tot+=b-dfs(i,b,0,num[i]);
	}
	   printf("%I64d\n",tot);
	}
	return 0;
}