HDU 1695 GCD 容斥

GCD

题目连接：

http://acm.hdu.edu.cn/showproblem.php?pid=1695

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

T
a b c d k

Output

ans

Sample Input

1
1 3 1 5 1

Sample Output

9

Hint

题意

问你gcd(i,j)=k有多少对，其中b>=i>=a,d>=j>=c
其中a和c恒等于1

题解：

很显然，我们知道一个结论，gcd(i,j)=k,b>=i>=a,d>=j>=c这个恒等于

gcd(i,j)=1,b/k>=i>=1,d/k>=j>=1

然后怎么办呢？我们暴力枚举每一个1<=i<=b/k的数，看在1<=j<=d/k里面有多少个和他互质的就好了

这个我们可以用容斥来做。

代码

#include<bits/stdc++.h>
using namespace std;
const int maxn = 2e5+5;
vector<int>pri[maxn];
void pre()
{
    for(int i=2;i<100007;i++)
    {
        int now = i;
        for(int j=2;j*j<=now;j++)
        {
            if(now%j==0)
            {
                pri[i].push_back(j);
                while(now%j==0)
                    now/=j;
            }
            if(now==1)break;
        }
        if(now>1)
            pri[i].push_back(now);
    }
}
int solve(int x,int tot)
{
    int res = 0;
    for(int i=1;i<(1<<pri[x].size());i++)
    {
        int num = 0;
        int tmp = 1;
        for(int j=0;j<pri[x].size();j++)
        {
            if((i>>j)&1)
            {
                num++;
                tmp*=pri[x][j];
            }
        }
        if(num%2==1)res+=tot/tmp;
        else res-=tot/tmp;
    }
    return tot-res;
}
int main()
{
    pre();
    int t;
    scanf("%d",&t);
    for(int cas=1;cas<=t;cas++)
    {
        int a,b,c,d,k;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        if(k==0)
        {
            printf("Case %d: 0
",cas);
            continue;
        }
        b/=k,d/=k;
        if(b<d)swap(b,d);
        long long ans = 0;
        for(int i=1;i<=b;i++)
            ans+=solve(i,min(i,d));
        printf("Case %d: %lld
",cas,ans);
    }
}