# BZOJ 2301 Problem b (莫比乌斯反演+容斥)

$F（n,m,k）$为满足$gcd(i,j)=k(1\leq i\leq n,1\leq j\leq m)$的对数。则$ans = F(b,d,k)-F(a-1,d,k)-F(c-1,b,k)+F(a-1,c-1,k)$

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxn=1e5+5;
bool vis[maxn];
int prime[maxn],mu[maxn];
int sum[maxn];
void init(){
memset(vis,false,sizeof(vis));
mu[1] = 1;
prime[0] = 0;
int cnt=0;
for(int i=2;i<maxn;++i){
if(!vis[i]){
mu[i] = -1;
sum[i] = 1;
prime[++cnt] = i;
}
for(int j=1;j<=cnt;++j){
if(i*prime[j] >= maxn)  break;
vis[i*prime[j]] = true;
if(i % prime[j]){
mu[i*prime[j]] = -mu[i];
sum[i*prime[j]] = mu[i] - sum[i];
}
else{
mu[i*prime[j]] = 0;
sum[i*prime[j]] = mu[i];
break;
}
}
}
for(int i =2;i<maxn;++i) sum[i]+=sum[i-1];
}
void prepare(){
int i,j,cnt=0;
mu[1]=sum[1]=1;
for(i=2;i<maxn;i++){
if(!vis[i])
prime[++cnt]=i,mu[i]=-1;
for(j=1;prime[j]*i<maxn;j++){
vis[prime[j]*i]=1;
if(i%prime[j]==0){
mu[prime[j]*i]=0;
break;
}
mu[prime[j]*i]=-mu[i];
}
sum[i]=sum[i-1]+mu[i];
}
}

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

int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt","r",stdin);
freopen("out.txt","w",stdout);
#endif
prepare();
LL a,b,c,d,k;
int T; scanf("%d",&T);
while(T--){
scanf("%lld %lld %lld %lld %lld",&a,&b,&c,&d,&k);
LL res=0;
res += gao(b,d,k);
res -= gao(a-1,d,k);
res -= gao(c-1,b,k);
res += gao(a-1,c-1,k);
printf("%lld\n",res);
}
return 0;
}
posted @ 2018-08-26 22:58  xiuwenL  阅读(...)  评论(...编辑  收藏