HDU 1695 GCD(容斥定理)

GCD

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11186    Accepted Submission(s): 4242

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:给两个集合，找出这两个集合中gcd为某个值的对数

思路:

b和d分别除以k之后的区间里面，只需要求gcd(x, y) = 1就可以了，这样子求出的数的对数不变。

这道题目还要求1-3 和 3-1 这种情况算成一种，因此只需要限制x

只需要枚举x，然后确定另一个区间里面有多少个y就可以了。因此问题转化成为区间（1， d）里面与x互素的数的个数

先求出x的所有质因数，因此（1，d）区间里面是x的质因数倍数的数都不会与x互素，因此，只需要求出这些数的个数，减掉就可以了。

如果w是x的素因子，则（1，d）中是w倍数的数共有d/w个。

#include 
#include 
#include 
using namespace std;

typedef long long ll;

const int MAXN=1e5+5;
int prime[MAXN][10];
int cnt[MAXN];

void sieve(int n)
{
    for(int i=2;i<=n;i++)
    {
        if(cnt[i]!=0) continue;
        prime[i][0]=i;
        cnt[i]=1;
        for(int j=2;j*i<=n;j++)
            prime[i*j][cnt[i*j]++]=i;
    }
}

ll solve(int m,int n,int pos)
{
    ll res=0;
    for(int i=1;i<(1<>=1) num+=j&1;
        int lcm=1;
        for(int j=0;j>j&1)
            {
                lcm=lcm/__gcd(lcm,prime[pos][j])*prime[pos][j];
                if(lcm>n) break;
            }
        }
        if(num%2==0) res-=(ll)(n/lcm);
        else res+=(ll)(n/lcm);
    }
    return res;
}

int main()
{
    sieve(100005);
    int t;
    scanf("%d",&t);
    int casi=1;
    while(t--)
    {
        int a,b,c,d,e;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&e);
        if(e==0)
        {
            printf("Case %d: %lld\n",casi++,0);
            continue;
        }
        b/=e;d/=e;
        if(b>d) swap(b,d);
        ll ans=0;
        for(int i=a;i<=b;i++)
        {
            ans+=(ll)d-(solve(cnt[i],d,i))-((ll)(i-1)-(solve(cnt[i],i-1,i)));
        }
        printf("Case %d: %lld\n",casi++,ans);
    }
    return 0;
}