hdu 1695 GCD （欧拉函数、容斥原理）

时间：2021-01-15 07:58:59

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


题目大意：求出[a,b]和[c,d]区间里面gcd（x,y）=k的数的对数。
思路：既然是求gcd为k的数的对数，最好还是先将b和d都除以k，这样问题就转化为[1,n]和[1,m]区间里面gcd（x,y）为1 的数的对数。由于题目里已经说明a和c 能够觉得是1，这样就更简单了。

对于一个[1,n]的区间。我们能够用欧拉函数算出总对数。

那么问题就能够分解成2个：
1、在[1,n]上用欧拉函数算出总对数。
2、在[n+1,m]上。计算在[1,n]里面的总对数，能够用容斥原理。
#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;

}


相关文章