这个莫比乌斯函数就是用来加速求gcd,lcm,约数之类问题的问题。对于区间[1,i]和[1,j]的里面gcd为k的个数,我们有朴素n^2logn算法,对于超过1e5的数据这种算法显然过于疲软,那么我们就需要线性处理,首先莫比乌斯函数是什么怎么选写多了肯定也就会了,我数学不好(假数学爱好者),还在努力搞懂,所以推导过程略过,作为计算机选手只用掌握套路就行,那么根据gcd的性质,gcd(i,j)== k显然有 gcd(i / k , j / k) == 1,所以常用套路是先用线性筛预处理出积性函数,使得n为m和n相对较小的那一个,枚举d从1到 n / k, 对于每个d 代表为gcd(i / k,j / k) == d对答案的贡献为
miu[d] * F(d * k)//F(x) = (n / x) * (n / x)为莫比乌斯函数
然后就行了,这么一来是O(n)的,完美
洛谷P1390 公约数的和
这个题第二个求和表达式不是从1开始的,没用分块,直接重新推的,减掉就行
#include <bits/stdc++.h>
using namespace std;
#define limit (2000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long int ll;
#define int ll
typedef unsigned long long ull;
inline ll read(){
ll sign = 1, x = 0;char s = getchar();
while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
return x * sign;
}//快读
void write(ll x){
if(x < 0) putchar('-'),x = -x;
if(x / 10) write(x / 10);
putchar(x % 10 + '0');
}
int prime[limit],tot,num[limit],phi[limit],miu[limit];
void get_prime(const int &n = 1e6){
memset(num,1,sizeof(num));
num[1] = num[0] = 0;
miu[1] = 1;
rep(i,2,n){
if(num[i])prime[++tot] = i,miu[i] = -1,phi[i] = i - 1;
for(int j = 1; j <= tot && prime[j] * i <= n ; ++j){
num[prime[j] * i] = 0;
if(i % prime[j] == 0){
miu[i * prime[j]] = 0;
break;
}else{
miu[i * prime[j]] = -miu[i];//莫比乌斯函数
}
}
}
}//素数筛
int n;
ll calc(int x){
ll ans = 0;
rep(d,1,n / x){
ans += miu[d] * (n / d / x) * (n / d / x);
}
return ans;
}
signed main() {
#ifdef LOCAL
FOPEN;
#endif
n = read();
get_prime(n);
ll ans = 0;
rep(i,1,n){
ans += i * calc(i);
}
ans -= ((n + 1) * n) / 2;
ans /= 2;
write(ans);
return 0;
}
遇到gcd == k的普通套路就是先降等号右边,然后莫比乌斯反演
洛谷P3455 ZAP-Queries
这个题问的是裸的莫反,易知此处选取莫比乌斯函数为F(d) = (n / d) * (m / d)
然后就套上跑出来即可,但是注意朴素解容易超时,需要用到除法值域分块,也不难的,求出来miu的前缀和就行
代码:
#include <bits/stdc++.h>
using namespace std;
#define limit (50000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long int ll;
typedef unsigned long long ull;
inline ll read(){
ll sign = 1, x = 0;char s = getchar();
while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
return x * sign;
}//快读
void write(ll x){
if(x < 0) putchar('-'),x = -x;
if(x / 10) write(x / 10);
putchar(x % 10 + '0');
}
int prime[limit],tot,num[limit],phi[limit],miu[limit];
void get_prime(const int &n = 1e6){
memset(num,1,sizeof(num));
num[1] = num[0] = 0;
miu[1] = 1;
rep(i,2,n){
if(num[i])prime[++tot] = i,miu[i] = -1,phi[i] = i - 1;
for(int j = 1; j <= tot && prime[j] * i <= n ; ++j){
num[prime[j] * i] = 0;
if(i % prime[j] == 0){
miu[i * prime[j]] = 0;
break;
}else{
miu[i * prime[j]] = -miu[i];//莫比乌斯函数
}
}
miu[i] += miu[i - 1];
}
}//素数筛
ll n,m,d;
ll F(ll x){
return (n / x) * (m / x);//莫比乌斯函数
}
ll calc(){
ll ans = 0;
for(int l = 1,r ; l <= min(n / d , m / d); l = r + 1){
//值域分块
ll t = n / d , s = m / d;
r = min(t / (t / l), s / (s / l));
ans += (miu[r] - miu[l - 1]) * (t / l) * (s / l);
}
return ans;
}
int main() {
#ifdef LOCAL
FOPEN;
#endif
int t = read();
get_prime(5e4 + 1);
while (t--){
n = read(),m = read(),d = read();
ll minn = min(n / d , m / d);
ll ans = calc();
printf("%lld\n",ans);
}
return 0;
}
然后如果下界不为1的,有上下界差分的,比如我区2018年区域赛的一道题,问两个区间有多少互质的数,可以转化为a-b和u-v的gcd==1的问题,来一遍莫反
莫比乌斯函数选取F(a,b,u,v) = ((b / k) - ((a - 1) / k)) * ((v / k) - ((u - 1)/ k))
然后理论上用值域分块即可,但不用好像也没啥关系(其实用欧拉函数和前缀和也行
Kattis/Cf Gym Coprime Integers:
#include <bits/stdc++.h>
using namespace std;
#define limit (10000000 + 5)//防止溢出
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define lowbit(i) i&(-i)//一步两步
#define EPS 1e-6
#define FASTIO ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long int ll;
typedef unsigned long long ull;
inline ll read(){
ll sign = 1, x = 0;char s = getchar();
while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
return x * sign;
}//快读
void write(ll x){
if(x < 0) putchar('-'),x = -x;
if(x / 10) write(x / 10);
putchar(x % 10 + '0');
}
int prime[limit],tot,num[limit],phi[limit],miu[limit];
void get_prime(const int &n = 1e7){
memset(num,1,sizeof(num));
num[1] = num[0] = 0;
miu[1] = 1;
rep(i,2,n){
if(num[i])prime[++tot] = i,miu[i] = -1,phi[i] = i - 1;
for(int j = 1; j <= tot && prime[j] * i <= n ; ++j){
num[prime[j] * i] = 0;
if(i % prime[j] == 0){
miu[i * prime[j]] = 0;
break;
}else{
miu[i * prime[j]] = -miu[i];//莫比乌斯函数
}
}
}
}//素数筛
ll d;
ll n,m,a,b;
ll F(ll k){
return ((n / k) - ((a - 1) / k)) * ((m / k) - ((b - 1)/ k));//莫比乌斯函数
}
ll calc(){
ll ans = 0;
for(ll k = 1; k <= min(n,m); ++k){
ans += miu[k] * F(k);//莫比乌斯反演
}
return ans;
}
int main() {
#ifdef LOCAL
FOPEN;
#endif
get_prime(1e7 + 1);
a = read(),n = read(),b = read(),m = read(),d =1;
ll ans = calc();
write(ans);
return 0;
}
完结撒花,具体问题肯定不可能是原模板,变形柿子怎么推是个问题
来源:oschina
链接:https://my.oschina.net/u/4385595/blog/4528261