【bzoj 3739】DZY loves math VIII - 乱搞数学题

　　103组询问，n≤107，求

∑i=1n∑j=1iμ(lcm(i,j)gcd(i,j))

　　说了这么多这题其实就是求∑i∑jμ(ij)吧。。。
　　首先注意到当gcd(i,j)>1时，若lcm(i,j)>1，则μ为0，无贡献，同时用lcm(i,j)=ij/gcd(i,j)展开，可以得到
　　

∑i=1n∑j=1i[gcd(i,j)==1]μ(ij)

　　因为gcd(i,j)==1，所以μ(ij)=μ(i)μ(j)，可得
　　

∑i=1n∑j=1i∑d∣i,d∣jμ(d)μ(i)μ(j)　　=∑i=1nμ(i)∑d∣iμ(d)∑j=1i/dμ(jd)

　　以下是玄学= =
　　打一个表会发现μ≠0的i的因子总数只有5∗107。。。
　　于是大力枚举有效的i然后枚举d，注意到∑i/dj=1μ(jd)总共只有O(n)个值，可以直接递推计算，于是最后一项可以直接加上去。
　　这样xjb搞一搞，差点没跑过去= =
　　时间复杂度O(能过)
　　233333333333333

/*    I will chase the meteor for you, a thousand times over.    Please wait for me, until I fade forever.    Just 'coz GEOTCBRL.*/#include <bits/stdc++.h>using namespace std;#define fore(i,u)  for (int i = head[u] ; i ; i = nxt[i])#define rep(i,a,b) for (int i = a , _ = b ; i <= _ ; i ++)#define per(i,a,b) for (int i = a , _ = b ; i >= _ ; i --)#define For(i,a,b) for (int i = a , _ = b ; i <  _ ; i ++)#define Dwn(i,a,b) for (int i = ((int) a) - 1 , _ = (b) ; i >= _ ; i --)#define fors(s0,s) for (int s0 = (s) , _S = s ; s0 ; s0 = (s0 - 1) & _S)#define foreach(it,s) for (__typeof(s.begin()) it = s.begin(); it != s.end(); it ++)#define mp make_pair#define pb push_back#define pii pair<int , int>#define fir first#define sec second#define MS(x,a) memset(x , (a) , sizeof (x))#define gprintf(...) fprintf(stderr , __VA_ARGS__)#define gout(x) std::cerr << #x << "=" << x << std::endl#define gout1(a,i) std::cerr << #a << '[' << i << "]=" << a[i] << std::endl#define gout2(a,i,j) std::cerr << #a << '[' << i << "][" << j << "]=" << a[i][j] << std::endl#define garr(a,l,r,tp) rep (__it , l , r) gprintf(tp"%c" , a[__it] , " \n"[__it == _])template <class T> inline void upmax(T&a , T b) { if (a < b) a = b ; }template <class T> inline void upmin(T&a , T b) { if (a > b) a = b ; }typedef long long ll;const int maxn = 10000003;const int maxm = 200007;const int mod = 1000000007;const int inf = 0x7fffffff;const double eps = 1e-7;typedef int arr[maxn];typedef int adj[maxm];inline int fcmp(double a , double b) {    if (fabs(a - b) <= eps) return 0;    if (a < b - eps) return -1;    return 1;}inline int add(int a , int b) { return ((ll) a + b) % mod ; }inline int mul(int a , int b) { return ((ll) a * b) % mod ; }inline int dec(int a , int b) { return add(a , mod - b % mod) ; }inline int Pow(int a , int b) {    int t = 1;    while (b) {        if (b & 1) t = mul(t , a);        a = mul(a , a) , b >>= 1;    }    return t;}#define gc getchar#define idg isdigit#define rd RD<int>#define rdll RD<long long>template <typename Type>inline Type RD() {    char c = getchar(); Type x = 0 , flag = 1;    while (!idg(c) && c != '-') c = getchar();    if (c == '-') flag = -1; else x = c - '0';    while (idg(c = gc()))x = x * 10 + c - '0';    return x * flag;}inline char rdch() {    char c = gc();    while (!isalpha(c)) c = gc();    return c;}#undef idg#undef gc// beginningarr vis , mu , v , S;int pr[maxn / 10] , ptt;int tot/* , cnt*/;int a[10001]/* , fr[100001]*/;ll ans[maxn];static int CLOCK;int x , res;void dfs(int p , int w) {    if (p == tot + 1) {        S[w] += x;        res += mu[w] * S[w];//      ++ CLOCK;        return;    }    dfs(p + 1 , w);    dfs(p + 1 , w * a[p]);}int N;inline void init() {    mu[1] = 1 , v[1] = 1 , ans[1] = 1;    rep (i , 2 , N) {        if (!vis[i])            pr[++ ptt] = i , mu[i] = -1 , v[i] = i;        rep (j , 1 , ptt) {            int t = i * pr[j];            if (t > N) break;            vis[t] = 1;            if (i % pr[j] == 0) { v[t] = v[i]; break;}            mu[t] = -mu[i] , v[t] = pr[j];        }    }    S[1] = 1;    rep (i , 2 , N) if (mu[i]) {        int t = i; tot = 0;        while (t > 1)            a[++ tot] = v[t] , t /= v[t];        res = 0;        x = mu[i];        dfs(1 , 1);        ans[i] = ans[i - 1] + res * x;    } else ans[i] = ans[i - 1];//  gout(CLOCK);}int n , m[1007];void input() {    n = rd();    rep (i , 1 , n) m[i] = rd() , upmax(N , m[i]);}void solve() {    init();    rep (i , 1 , n) printf("%lld\n" , ans[m[i]]);}int main() {    #ifndef ONLINE_JUDGE        freopen("data.txt" , "r" , stdin);    #endif    input();    solve();    return 0;}