# [BZOJ2301][HAOI2011]Problem b

BZOJ
Luogu
Description

Input

Output

Sample Input
2
2 5 1 5 1
1 5 1 5 2
Sample Output
14
3
HINT
100%的数据满足：1≤n≤50000，1≤a≤b≤50000，1≤c≤d≤50000，1≤k≤50000

# sol

## code

#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
const int N = 50005;
const int n = 50000;
int gi()
{
int x=0,w=1;char ch=getchar();
while ((ch<'0'||ch>'9')&&ch!='-') ch=getchar();
if (ch=='-') w=0,ch=getchar();
while (ch>='0'&&ch<='9') x=(x<<3)+(x<<1)+ch-'0',ch=getchar();
return w?x:-x;
}
int mu[N],pri[N],tot,s[N];
bool zhi[N];
void Mobius()
{
zhi[1]=true;mu[1]=1;
for (int i=2;i<=n;i++)
{
if (!zhi[i]) pri[++tot]=i,mu[i]=-1;
for (int j=1;j<=tot&&i*pri[j]<=n;j++)
{
zhi[i*pri[j]]=1;
if (i%pri[j]) mu[i*pri[j]]=-mu[i];
else {mu[i*pri[j]]=0;break;}
}
}
for (int i=1;i<=n;i++) s[i]=s[i-1]+mu[i];
}
ll calc(int a,int b,int k)
{
a/=k;b/=k;
if (a>b) swap(a,b);
int i=1,j;ll ans=0;
while (i<=a)
{
j=min(a/(a/i),b/(b/i));
ans+=1ll*(s[j]-s[i-1])*(a/i)*(b/i);
i=j+1;
}
return ans;
}
int main()
{
int T=gi();
Mobius();
while (T--)
{
int a=gi()-1,b=gi(),c=gi()-1,d=gi(),k=gi();
printf("%lld\n",calc(b,d,k)-calc(a,d,k)-calc(b,c,k)+calc(a,c,k));
}
return 0;
}
posted @ 2018-01-04 09:56  租酥雨  阅读(...)  评论(...编辑  收藏