bzoj 2301: [HAOI2011]Problem b 莫比乌斯反演

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define esp 0.00000000001
#define pi 4*atan(1)
const int N=1e5+10,M=1e7+10,inf=1e9+10,mod=1e9+7;
int mu[N], p[N], np[N], cnt, sum[N];
void init() {
    mu[1]=1;
    for(int i=2; i<N; ++i) {
        if(!np[i]) p[++cnt]=i, mu[i]=-1;
        for(int j=1; j<=cnt && i*p[j]<N; ++j) {
            int t=i*p[j];
            np[t]=1;
            if(i%p[j]==0) { mu[t]=0; break; }
            mu[t]=-mu[i];
        }
    }
    for(int i=1;i<N;i++)
    sum[i]=sum[i-1]+mu[i];
}
ll getans(int b,int d)
{
    if(b>d)swap(b,d);
    ll ans=0;
    for(int L=1,R=0;L<=b;L=R+1)
    {
        R=min(b/(b/L),d/(d/L));
        ans+=(ll)(sum[R]-sum[L-1])*(b/L)*(d/L);
    }
    return ans;
}
int main()
{
    int T;
    init();
    scanf("%d",&T);
    while(T--)
    {
        int a,c,b,d,k;
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        printf("%lld\n",getans(b/k,d/k)+getans((a-1)/k,(c-1)/k)-getans((a-1)/k,d/k)-getans(b/k,(c-1)/k));
    }
    return 0;
}