Bzoj 3813 Odd Country Euler function + segment tree + multiplicative inverse

The main idea: to give a sequence, support the modification operation, the sequence of the series of the product of the Euler function. The maximum quality factor for each number does not exceed 281.

Idea:φ (n) = n * (1-1/p1) * (1-1/p2) * (1-1/p3) * (1-1/P4) ... * (1-1/PN)

= N/(P1 * p2 * p3 * ... * pn) * ((p1-1) * (p2-1) * (p3-1) * ... * (pn-1))

So this thing just need to maintain the index in the interval, with bitset maintenance of factorization, set out to use the multiplication inverse, no.

CODE:

#include <bitset> #include <cstdio> #include <cstring> #include <iostream> #include < algorithm> #define P 19961993#define MAX 100010using namespace std; #define LEFT (POS << 1) #define RIGHT (POS <& Lt 1|1) #define MAX (a) > (b)?          (a):(B)) struct ask{int flag,x,y;    void Read () {scanf ("%d%d%d", &flag,&x,&y);  }}ask[max];    struct segtree{bitset<61> Prime;          int total; Segtree (bitset<61> _,int __):p Rime (_), total (__) {} segtree () {} Segtree operator + (const segtree &a) const    {return Segtree (Prime|a.prime, (long Long) total * a.total% P);  }}tree[max << 2];  int G[300],g[300];long long Inv[300];int asks,cnt;    inline bool Judge (int x) {for (int i = 2; I * i <= x; ++i) if (x% i = = 0) return false;  return true;}    void Shake () {inv[1] = 1;  for (int i = 2; I <= 281; ++i) inv[i] = (P/i) * inv[p% i]% P;}  void Pretreatment () {  Shake ();    int cnt = 0;  for (int i = 2; I <= 281; ++i) if (Judge (i)) g[i] = ++cnt,g[cnt] = i;}        void Buildtree (int l,int r,int POS) {if (L = = r) {Tree[pos].prime[g[3]] = 1;        Tree[pos].total = 3;    return;    } int mid = (L + r) >> 1;    Buildtree (L,mid,left);    Buildtree (mid + 1,r,right); Tree[pos] = Tree[left] + tree[right];    } segtree Ask (int l,int r,int x,int y,int pos) {if (L = = x && r = = y) return tree[pos];    int mid = (L + r) >> 1;    if (y <= mid) return Ask (L,mid,x,y,left);    if (x > Mid) return Ask (mid + 1,r,x,y,right);    Segtree left = Ask (L,mid,x,mid,left);    Segtree right = Ask (mid + 1,r,mid + 1,y,right);  return left + right;}        void Modify (int l,int r,int x,int pos,segtree c) {if (L = = r) {Tree[pos] = C;    return;    } int mid = (L + r) >> 1;    if (x <= mid) Modify (L,MID,X,LEFT,C);    else Modify (mid + 1,r,x,right,c); Tree[pos] = Tree[left] + treE[right];}    int main () {pretreatment ();    Cin >> asks; for (int i = 1; I <= asks; ++i) Ask[i].    Read ();    CNT = MAX-1;    Buildtree (1,cnt,1);        for (int flag,x,y,i = 1; I <= asks; ++i) {flag = Ask[i].flag;        x = Ask[i].x,y = Ask[i].y;            if (flag) {bitset<61> p;            for (int j = 1; J <=; ++j) if (y% g[j] = = 0) P[j] = 1;            Segtree c (p,y);        Modify (1,CNT,X,1,C);            } else {Segtree ans = Ask (1,cnt,x,y,1); for (int j = 1; J <=; ++j) if (Ans.prime[j]) {ans.total = (long long) ans.total * I                    NV[G[J]])% P;                Ans.total = ((Long Long) ans.total * (G[j]-1))% P;        } printf ("%d\n", ans.total); }} return 0;}

Bzoj 3813 Odd Country Euler function + segment tree + multiplicative inverse