Codeforces 547C Mike and Foam

Codeforces 547C Mike and Foam

 

Question:

Number of n and q operations

N number a1-an below

 

There is an empty set at the beginning.

Each operation has a number u (1 <= u <= n), which means to insert a [u] into the Set (if there is no a [u] in the set). otherwise, a [u] will be deleted from the set.

After each operation is completed, the number of logarithm numbers in the current set is output to each other.

 

Ideas:

Break down the quality factor, and then let us look at the number of non-mutual quality with a [u]. Just make a difference.

 

PS: cf ~

 

 

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             Using namespace std; template
            
              Inline bool rd (T & ret) {char c; int sgn; if (c = getchar (), c = EOF) return 0; while (c! = '-' & (C <'0' | c> '9') c = getchar (); sgn = (c = '-')? -1: 1; ret = (c = '-')? 0: (c-'0'); while (c = getchar (), c> = '0' & c <= '9 ') ret = ret * 10 + (c-'0'); ret * = sgn; return 1;} template
             
               Inline void pt (T x) {if (x <0) {putchar ('-'); x =-x;} if (x> 9) pt (x/10); putchar (x % 10 + '0');} typedef long ll; const int N = 5e5 + 10; int prime [N], primenum; // math has primenum prime numbers. h void PRIME (ll Max_Prime) {primenum = 0; prime [primenum ++] = 2; for (ll I = 3; I <= Max_Prime; I + = 2) for (ll j = 0; j
              
                Sqrt (double) I) | j = primenum-1) {prime [primenum ++] = I; break ;}} int n, q; int a [N]; vector
               
                 G [N]; ll ans; void pre (int u, int x) {for (int I = 0; prime [I] * prime [I] <= x; I ++) {if (x % prime [I] = 0) {while (x % prime [I] = 0) x/= prime [I]; G [u]. push_back (prime [I]) ;}} if (x! = 1) G [u]. push_back (x);} set
                
                  S; int f [N]; int c [N], top; void dfs (int t, int x, int cnt, int flag, int val) {if (t> = (int) G [x]. size () {if (cnt = 0) return; ans + = f [val] * flag; c [top ++] = val; return ;} dfs (t + 1, x, cnt, flag, val); dfs (t + 1, x, cnt + 1, flag *-1, val * G [x] [t]);} void p () {printf (set :); for (auto it: s) {printf (% d, it );} puts () ;}int one; int main () {PRIME (N); rd (n); rd (q); for (int I = 1; I <= n; I ++) {rd (a [I]); pre (I, a [I]);} ans = 0; one = 0; while (q --) {int u; rd (u); if (a [u] = 1) {if (s. count (u) {one --; s. erase (u); ans-= s. size ();} else {ans + = s. size (); one ++; s. insert (u) ;}} else {// printf (ans: % d, one: % d); p (); if (s. count (u) {s. erase (u); ans-= s. size (); top = 0; dfs (0, u, 0,-1, 1); ans --; // Delete the combination of yourself and yourself for (int I = 0; I <top; I ++) f [c [I] --;} else {ans + = s. size (); s. insert (u); top = 0; dfs (0, u, 0, 1, 1); for (int I = 0; I <top; I ++) f [c [I] ++ ;}} cout <; pt (ans); puts () ;}return 0 ;}