P2522 [HAOI2011]Problem b

P2522 [HAOI2011]Problem b（我的第一道莫比乌斯反演）

 题解：

根据题意写出函数表达式：

(fleft ( k ight )=sum_{i=1}^{n}sum_{j=1}^{m}left [ gcdleft ( i,j ight)=k ight ])　　表示(1leq ileq n,1leq jleq m) ， (gcdleft ( i,j ight)=k)成立的(left ( i,j ight ))对数

令(Fleft ( x ight )=sum_{x|d}fleft ( d ight ))，代表 (gcd=d且是x的倍数的left ( i,j ight )对数)；

我们知道(x|gcdleft ( i,j ight)=left ( x|i ight )wedge left ( x|j ight ))，又因为(1sim n之间整除x的数有left lfloor frac{n}{x} ight floor个)，(1sim m之间整除x的数有left lfloor frac{m}{x} ight floor个)

所以(Fleft ( x ight )=left lfloor frac{n}{x} ight floorcdot left lfloor frac{m}{x} ight floor)

于是：((Fleft ( x ight )=sum_{x|d}^{d}=left lfloor frac{n}{x} ight floorcdot left lfloor frac{m}{x} ight floor)

根据第二种莫比乌斯反演：则(fleft ( x ight )=sum_{x|d}mu left ( frac{d}{x} ight )Fleft ( d ight ))

带入(Fleft ( d ight )=left lfloor frac{n}{d} ight floorcdot left lfloor frac{m}{d} ight floor)　得

(fleft ( x ight )=sum_{x|d}mu left ( frac{d}{x} ight )left lfloor frac{n}{d} ight floorleft lfloor frac{m}{d} ight floor)

此时(x=d)，(dleq minleft ( n，m ight ))，则(fleft ( x ight )=sum_{d=1}^{minleft ( n,m ight )}mu left ( frac{d}{x} ight )left lfloor frac{n}{d} ight floorleft lfloor frac{m}{d} ight floor)

令(t=frac{d}{x})，则(d=xt)；(dleq minleft ( n，m ight ))，则(t=frac{d}{x}leq minleft ( frac{n}{x},frac{m}{x} ight ))

于是：(fleft ( d ight )=sum_{t=1}^{minleft ( frac{n}{d},frac{m}{d} ight )}mu left ( t ight )left lfloor frac{n}{td} ight floorleft lfloor frac{m}{td} ight floor)

(d)是题目所给的常数，这里是(k)；

注意题目不是(1leq ileq b,1leq jleq d)，所以还要用到容斥原理，算出(1sim b,1sim d)，减去(1sim a,1sim d)，再减去(1sim c,1sim b)，再加上(1sim a,1sim c)

用整除分块+前缀和（预处理莫比乌斯函数）

AC_Code:

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
const int maxn = 5e4+10;
const int mod = 2333333;
const int inf = 0x3f3f3f3f;
#define gok ios::sync_with_stdio(false);cin.tie(0);cout.tie(0)
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
#define rep(i,first,second) for(ll i=first;i<=second;i++)
#define dep(i,first,second) for(ll i=first;i>=second;i--)
#define erep(i,u) for(ll i=head[u];~i;i=e[i].nxt)

int a,b,c,d,k;
bool prime[maxn];
int Prime[maxn],tot,mu[maxn],sum[maxn];
void get_mu(){
    memset(prime,true,sizeof(prime));
    prime[0]=prime[1]=false;
    mu[1]=1;
    for(int i=2;i<=50000;i++){
        if( prime[i] ) Prime[++tot]=i,mu[i]=-1;
        for(int j=1;j<=tot&&i*Prime[j]<=50000;j++){
            prime[i*Prime[j]]=false;
            if( i%Prime[j]==0 ){
                mu[i*Prime[j]]=0;
                break;
            }
            mu[i*Prime[j]]=-mu[i];
        }
    }
    for(int i=1;i<=50000;i++) sum[i]=sum[i-1]+mu[i];
}

int cal(int n,int m){
    int res=0;
    for(int i=1,j;i<=min(n,m);i=j+1){
        j=min(n/(n/i),m/(m/i));
        res += (sum[j]-sum[i-1])*(n/i)*(m/i);
    }
    return res;
}

int main()
{
    int t;
    scanf("%d",&t);
    get_mu();
//    rep(i,1,10) cout<<mu[i]<<' '<<sum[i]<<endl;
    while( t-- ){
        scanf("%d%d%d%d%d",&a,&b,&c,&d,&k);
        printf("%d
",cal(b/k,d/k)-
               cal((a-1)/k,d/k)-
               cal(b/k,(c-1)/k)+
               cal((a-1)/k,(c-1)/k));
    }
    return 0;
}

 再放一道例题：注意下边界都是从1开始，且（a,b）（b,a）算一种，注意去重

AC_Code:

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
const int mod = 1e9+7;
const int inf = 0x3f3f3f3f;
#define rep(i,first,second) for(ll i=first;i<=second;i++)
#define dep(i,first,second) for(ll i=first;i>=second;i--)

bool prime[maxn];
ll Prime[maxn],tot,mu[maxn];
ll ans,sum[maxn];
ll a,b,c,d,k;

void get_mu(){
    memset(prime,true,sizeof(prime));
    prime[0]=prime[1]=false;
    mu[1]=1;
    rep(i,2,100000){
        if( prime[i] ) Prime[++tot]=i,mu[i]=-1;
        for(ll j=1;j<=tot&&i*Prime[j]<=100000;j++){
            prime[i*Prime[j]]=false;
            if( i%Prime[j]==0 ){
                mu[i*Prime[j]]=0;
                break;
            }
            mu[i*Prime[j]]=-mu[i];
        }
    }
    rep(i,1,100000) sum[i]=sum[i-1]+mu[i];
}

ll cal(ll x,ll y){
    ll up=min(x,y);
    ll res=0;
    for(ll i=1,j;i<=up;i=j+1){
        j=min(x/(x/i),y/(y/i));
        res += (sum[j]-sum[i-1])*(x/i)*(y/i);
    }
    return res;
}

int main()
{
    int t,cas=0;
    scanf("%d",&t);
    get_mu();
    while( t-- ){
        scanf("%lld%lld%lld%lld%lld",&a,&b,&c,&d,&k);
        if( k==0 ){
            printf("Case %d: 0
",++cas);
            continue;
        }
        b=b/k;d=d/k;
        ll up=min(b,d);
        ll ans1=cal(b,d);//重复区有两倍,要减去一倍
        ll ans2=cal(up,up);//重复区的那两倍数算出来
        printf("Case %d: %lld
",++cas,ans1-ans2/2);//重复区的那两倍减去其中一倍即可
    }
    return 0;
}