HDU 1695 GCD (欧拉函数，容斥原理)

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 9046    Accepted Submission(s): 3351

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

对于求x在1~n之间，y在1~m之间的gcd(x,y)=k;
就相当于求x在1~n/k之间，y在1~m/k之间的gcd(x,y)=1;即x,y互质的对数
对于欧拉函数，可以求比n小的和n互质的个数。
而容斥原理可以求1~指定范围，和n互质的个数。
所以我们枚举一个区间的数，然后求这个数在另一个区间的互质的个数。
容斥原理可以解决，但是为了学习熟悉欧拉函数，所以可以分成两段，一段用欧拉函数，另一段用容斥原理。

求解欧拉函数，可以用线性素数晒求解，这样同时打了一个素数表，为容斥原理服务

#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <stdio.h>
#include <math.h>
#include <bitset>

using namespace std;
typedef long long int LL;
#define MAX 1000000
bool check[MAX+5];
LL fai[MAX+5];
LL prime[MAX+5];
LL sprime[MAX+5];
LL q[MAX+5];
int cnt;
void eular()//线性筛求解欧拉函数
{
	memset(check,false,sizeof(check));
	fai[1]=1;
	int tot=0;
	for(int i=2;i<=MAX+5;i++)
	{
		if(!check[i])
		{
            prime[tot++]=i;
			fai[i]=i-1;
		}
		for(int j=0;j<tot;j++)
		{
			if(i*prime[j]>MAX+5) break;
			check[i*prime[j]]=true;
			if(i%prime[j]==0)
			{
				fai[i*prime[j]]=fai[i]*prime[j];
				break;
			}
			else
			{
				fai[i*prime[j]]=fai[i]*(prime[j]-1);
			}
		}
	}
}
void Divide(LL n)//分解质因子
{
    cnt=0;
    LL t=(LL)sqrt(1.0*n);
    for(LL i=0; prime[i]<=t; i++) {
        if(n%prime[i]==0) {
            sprime[cnt++]=prime[i];
            while(n%prime[i]==0)
                n/=prime[i];
        }
    }
    if(n>1)
        sprime[cnt++]=n;
}
LL Ex(LL n)//容斥原理之队列实现
{
    
    LL sum=0;
    LL t=1;
    q[0]=-1;
    for(LL i=0; i<cnt; i++) {
        LL x=t;
        for(LL j=0; j<x; j++){
            q[t]=q[j]*sprime[i]*(-1);
            t++;
        }
    }
    for(LL i=1; i<t; i++)
        sum+=n/q[i];
    return sum;
}
int main()
{
	int t;
	scanf("%d",&t);
	eular();
	int cas=0;
	int a,b,c,d,k;
	while(t--)
	{
		scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
		if(k==0||k>b||k>d)
		{
			printf("Case %d: 0
",++cas);
			continue;
		}
		if(b>d) swap(b,d);
		b/=k;d/=k;
		LL ans=0;
		for(int i=1;i<=b;i++)
             ans+=fai[i];
		for(int i=b+1;i<=d;i++)
		{ Divide(i);ans+=(b-Ex(b));}
		printf("Case %d: %lld
",++cas,ans);
	}
	return 0;

}