#4320. ShangHai2006 Homework

根号分治

(f_x) 表示询问为 (x) 的答案，这一部分预处理的复杂度是 (n sqrt n)

(pt_i) 表示距离 (i) 最近的数

(tag_i) 表示距离 (i) 块最近的数

然后就可以做到 (sqrt) 修改， (1) ~ (sqrt) 查询了

// powered by c++11
// by Isaunoya
#include <bits/stdc++.h>
#define rep(i, x, y) for (register int i = (x); i <= (y); ++i)
#define Rep(i, x, y) for (register int i = (x); i >= (y); --i)
using namespace std;
using db = double;
using ll = long long;
using uint = unsigned int;
#define Tp template
using pii = pair<int, int>;
#define fir first
#define sec second
Tp<class T> void cmax(T& x, const T& y) {if (x < y) x = y;} Tp<class T> void cmin(T& x, const T& y) {if (x > y) x = y;}
#define all(v) v.begin(), v.end()
#define sz(v) ((int)v.size())
#define pb emplace_back
Tp<class T> void sort(vector<T>& v) { sort(all(v)); } Tp<class T> void reverse(vector<T>& v) { reverse(all(v)); }
Tp<class T> void unique(vector<T>& v) { sort(all(v)), v.erase(unique(all(v)), v.end()); }
const int SZ = 1 << 23 | 233;
struct FILEIN { char qwq[SZ], *S = qwq, *T = qwq, ch;
#ifdef __WIN64
#define GETC getchar
#else
  char GETC() { return (S == T) && (T = (S = qwq) + fread(qwq, 1, SZ, stdin), S == T) ? EOF : *S++; }
#endif
  FILEIN& operator>>(char& c) {while (isspace(c = GETC()));return *this;}
  FILEIN& operator>>(string& s) {while (isspace(ch = GETC())); s = ch;while (!isspace(ch = GETC())) s += ch;return *this;}
  Tp<class T> void read(T& x) { bool sign = 0;while ((ch = GETC()) < 48) sign ^= (ch == 45); x = (ch ^ 48);
    while ((ch = GETC()) > 47) x = (x << 1) + (x << 3) + (ch ^ 48); x = sign ? -x : x;
  }FILEIN& operator>>(int& x) { return read(x), *this; } FILEIN& operator>>(ll& x) { return read(x), *this; }
} in;
struct FILEOUT {const static int LIMIT = 1 << 22 ;char quq[SZ], ST[233];int sz, O;
  ~FILEOUT() { flush() ; }void flush() {fwrite(quq, 1, O, stdout); fflush(stdout);O = 0;}
  FILEOUT& operator<<(char c) {return quq[O++] = c, *this;}
  FILEOUT& operator<<(string str) {if (O > LIMIT) flush();for (char c : str) quq[O++] = c;return *this;}
  Tp<class T> void write(T x) {if (O > LIMIT) flush();if (x < 0) {quq[O++] = 45;x = -x;}
		do {ST[++sz] = x % 10 ^ 48;x /= 10;} while (x);while (sz) quq[O++] = ST[sz--];
  }FILEOUT& operator<<(int x) { return write(x), *this; } FILEOUT& operator<<(ll x) { return write(x), *this; }
} out;
#define int long long
const int maxn = 3e5 ;
const int S = 550 ;
const int inf = 1e9 ;
int f[S + 5] , tag[S + 5] , pt[maxn + 5] , bl[maxn + 5] ;
void add(int x) {
	for(int i = 1 ; i <= S ; i ++) 
		cmin(f[i] , x % i) ;
	cmin(tag[bl[x]] , x) ; cmin(pt[x] , x) ;
	for(int i = x - 1 ; i >= (bl[x] - 1) * S ; i --)
		cmin(pt[i] , pt[i + 1]) ;
	for(int i = bl[x] - 1 ; i ; i --)
		cmin(tag[i] , tag[i + 1]) ;
}

int qwq(int x) {
	return min(pt[x] , tag[bl[x] + 1]) ;
}

int qry(int x) {
	if(x <= S) return f[x] ;
	int ans = qwq(1) % x ;
	for(int i = x ; i <= maxn ; i += x) {
		int t = qwq(i) - i ;
		if(t < x) cmin(ans , t) ;
	}
	return ans ;
}

signed main() {
  // code begin.	
	memset(f , 0x3f , sizeof(f)) ;
	memset(tag , 0x3f , sizeof(tag)) ;
	memset(pt , 0x3f , sizeof(pt)) ;
	int q ;
	in >> q ;
	rep(i , 1 , maxn) bl[i] = (i - 1) / S + 1 ;
	while(q --) {
		char opt ;
		in >> opt ;
		int x ;
		in >> x ;
		if(opt == 'A') 
			add(x) ;
		else
			out << qry(x) << '
' ;
	}
	return 0;
  // code end.
}