HDU1695（容斥原理）

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 6017    Accepted Submission(s): 2203

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).

题意：在[1，b]和[1，d]中各选一个数i,j,是的GCD(i,j)=k,求满足的（i,j）对数，(i,j)和(j,i)算一种

思路：等价于求[1,b/k]和[1,d/k]中各选一个数i,j，使得i，j互素

对于i>=j，只需要累加小于i且与i互质的数的个数就好了，这个可以用欧拉函数求

对于i

不互素的个数，然后用i减去这部分就是互素的个数了

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
using namespace std;

typedef __int64 ll;
const int MAXN = 100010;
int isprime[MAXN];
int prime[MAXN];
int cnt;

void getprime()
{
    cnt=0;
    isprime[1]=1;
    for(int i=2;i<=100000;i++)if(!isprime[i]){
        prime[cnt++]=i;
        if(i>350)continue;
        for(int j=i*i;j<=100000;j+=i){
            isprime[j]=1;
        }
    }
}

int euler[MAXN];

void geteuler()
{
    euler[1]=1;
    for(int i=2;i<=100000;i++)if(!euler[i]){
        for(int j=i;j<=100000;j+=i){
            if(!euler[j])euler[j]=j;
            euler[j]=euler[j]/i*(i-1);
        }
    }
}

int calc(int n,int m)
{
    int stk[30];
    int sz=0;
    int mx=(int)sqrt(m+0.5);
    for(int i=0;prime[i]<=mx;i++){
        if(m%prime[i]==0)stk[sz++]=prime[i];
        while(m%prime[i]==0)m/=prime[i];
    }
    if(m!=1)stk[sz++]=m;
    int ans=0;
    for(int s=1;s<(1