hdu 1695 综合数论 欧拉函数 分解质因子 容斥原理 打印素数表 帅呆了的一个题目 详解

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3112    Accepted Submission(s): 1095

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

 

Recommend

wangye

/*
题目大意：

给你a, b, c, d, k; 找出这样的一队 x, y， 使得 gcd(x , y) = k, 并且x ∈[1, b], y ∈ [1, d], 问有多少对符合要求的(x, y)。

------------------------------------------------------------------------------


思路: gcd(x, y) == k 说明x,y都能被k整除, 但是能被k整除的未必gcd=k , 必须还要满足互质关系. 问题就转化为了求1~a/k 和 1~b/k间互质对数的问题可以把a设置为小的那个数,
那么以y>x来保持唯一性(题目要求, 比如[1,3] =[3,1] )

接下来份两种情况:

1. y <= a , 那么对数就是 1~a的欧拉函数的累计和(容易想到)

2. y >= a , 这个时候欧拉函数不能用了,怎么做?　　可以用容斥原理,把y与1~a互质对数问题转换为y的质数因子在1~a范围内能整除的个数(质数分解和容斥关系)，

*/

#include <iostream>
#include <stdio.h>
#include <memory.h>
#include<math.h>
#include <vector>
using namespace std;

const int N = 100005;
typedef long long LL;
#define maxn 100005
LL phi[N];
vector<LL> link[N];
int vis[1000000+5],c;  
LL prime[79000];  
void get_prime()  //打印素数表模板
{  
    int i,j,n,m;  
    c=0;  
    n=1000000;  
    m=(int)sqrt(n+0.5);  
    memset(vis,0,sizeof(vis));  
    for(i=2;i<=m;i++)   
        if(!vis[i])  
        {  
            for(j=i*i;j<=n;j+=i)  
                vis[j]=1;  
        }  
    for(i=2;i<=n;i++) if(!vis[i])  
        prime[c++]=i;  
}  


void get_PHI()  //模板  得到1->n之内 与n互质的数的个数  存入phi[n]
{  
    int i,j;  
    for (i = 1; i <= maxn; i++) phi[i] = i;  
    for (i = 2; i <= maxn; i += 2) phi[i] /= 2;  
    for (i = 3; i <= maxn; i += 2) if(phi[i] == i)  
    {  
        for (j = i; j <= maxn; j += i)  
            phi[j] = phi[j] / i * (i - 1);  
    }  
}

void init()     //求每一个数的质因数，vector储存  
{
    LL i, j, k;
    for(i = 1; i < N; i++)//求n的质因数  也是模板
    {
        k = i;
        for(j = 0; prime[j]*prime[j] <= k; j++)
        {
            if(k%prime[j] == 0)
            {
                link[i].push_back(prime[j]);
                while(k%prime[j] == 0)
                    k /= prime[j];
            }
            if(k == 1) break;
        }
        if(k > 1) link[i].push_back(k);
    }
}

LL make_ans(LL num,LL n)//1到num中的所有数与n的m个质因子不互质的数的个数 注意是不互质哦    容斥原理
{  
    LL ans=0,tmp,i,j,flag;  
	for(i=1;i<(LL)(1<<link[n].size());i++)  
    { //用二进制来1,0来表示第几个素因子是否被用到,如m=3，三个因子是2,3,5，则i=3时二进制是011，表示第2、3个因子被用到    
        tmp=1,flag=0;  
        for(j=0;j<link[n].size();j++)   
            if(i&((LL)(1<<j)))//判断第几个因子目前被用到   
                flag++,tmp*=link[n][j];//第j个质因子link[n][j]
        if(flag&1)//容斥原理，奇加偶减    
            ans+=num/tmp;  
        else  
            ans-=num/tmp;  
    }  
    return ans;  
}  


int main()
{
    LL i, a, b, c, d, k, sum, t, zz = 1;//longlong型的数据 可以用%I64d 来输入输出
    get_prime();
	get_PHI();
    init();
    scanf("%I64d", &t);
    while(t--)
    {
        scanf("%I64d %I64d %I64d %I64d %I64d", &a, &b, &c, &d, &k);
        if(k == 0 || k > b || k > d)
        {
            printf("Case %I64d: 0\n", zz++);
            continue;
        }
        if(b > d) swap(b, d);//保持d较大
        b /= k;
        d /= k;
        sum = 0;
        for(i = 1; i <= b; i++)
        {
            sum += phi[i];
        }
        for(i = b+1; i <= d; i++)
        {
            sum += b - make_ans(b, i);
        }
        printf("Case %I64d: %I64d\n", zz++, sum);
    }

    return 0;
}