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\n",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\n",cas,ans);
}
}