# bzoj 2301 莫比乌斯反演

/** @Date    : 2017-09-09 19:21:18
* @FileName: bzoj 2301 莫比乌斯反演 多组 范围内 GCD=k.cpp
* @Platform: Windows
* @Author  : Lweleth (SoungEarlf@gmail.com)
* @Version : $Id$
*/
#include <bits/stdc++.h>
#define LL long long
#define PII pair<int ,int>
#define MP(x, y) make_pair((x),(y))
#define fi first
#define se second
#define PB(x) push_back((x))
#define MMG(x) memset((x), -1,sizeof(x))
#define MMF(x) memset((x),0,sizeof(x))
#define MMI(x) memset((x), INF, sizeof(x))
using namespace std;

const int INF = 0x3f3f3f3f;
const int N = 5e4+20;
const double eps = 1e-8;

int pri[N];
int mu[N];
int c = 0;
int vis[N];
LL sum[N];

void prime()
{
MMF(vis);
mu[1] = 1;
for(int i = 2; i < N; i++)
{
if(!vis[i])
pri[c++] = i, mu[i] = -1;
for(int j = 0; j < c && i * pri[j] < N; j++)
{
vis[i * pri[j]] = 1;
if(i % pri[j] == 0)
{
mu[i * pri[j]] = 0;
break;
}
else mu[i * pri[j]] = -mu[i];
}
}
sum[0] = 0;
for(int i = 1; i < N; i++)
sum[i] += sum[i - 1] + mu[i];
}

LL getsum(LL n, LL m)
{
LL ans = 0;
if(n > m) swap(n, m);
for(int i = 1, last; i <= n; i = last + 1)
{
last = min(n/(n/i), m/(m/i));
ans += (LL)(n / i) * (m / i) * (sum[last] - sum[i - 1]);
}
return ans;
}

int main()
{
int T;
prime();
cin >> T;
while(T--)
{
LL a, b, c, d, k;
scanf("%lld%lld%lld%lld%lld", &a, &b, &c, &d, &k);
a = (a - 1) / k;
b = b / k;
c = (c - 1) / k;
d = d / k;
LL ans = getsum(a, c) + getsum(b, d) - getsum(a, d) - getsum(b, c);//容斥
printf("%lld\n", ans);
}
return 0;
}

posted @ 2017-09-12 21:31  Lweleth  阅读(...)  评论(...编辑  收藏