#6039. 「雅礼集训 2017 Day5」珠宝 [决策单调性]

满足决策单调性
枚举增量，然后枚举余数，就可以了QAQ

// 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
int n , m , mx = 0 ;
vector < int > w[333] ;
const int maxn = 5e4 + 45 ;
int dp[maxn] , g[maxn] , f[maxn] ;
int now ;
void solve(int l , int r , int x , int y) {
	if(l > r) return ; int mid = l + r >> 1 ;
	f[mid] = g[mid] ; int p = mid ;
	for(int i = x ; i <= y && i < mid ; i ++) {
		if(mid - i > w[now].size()) continue ;
		int k = g[i] + w[now][mid - i - 1] ;
		if(k > f[mid]) f[mid] = k , p = i ;
	}
	solve(l , mid - 1 , x , p) ;
	solve(mid + 1 , r , p , y) ;
}
signed main() {
  // code begin.
	in >> n >> m ;
	rep(i , 1 , n) {
		int c , v ; in >> c >> v ;
		w[c].pb(v) ; cmax(mx , c) ;
	}
	rep(i , 1 , mx) {
		if(! sz(w[i])) continue ;
		sort(w[i]) , reverse(w[i]) ;
		for(int j = 1 ; j < sz(w[i]) ; j ++)
			w[i][j] += w[i][j - 1] ;
	}
	for(int i = mx ; i ; i --) {
		if(! sz(w[i])) continue ;
		now = i ;
		for(int j = 0 ; j < i ; j ++) {
			int tot = 0 ;
			for(int k = j ; k <= m ; k += i)
				g[++ tot] = dp[k] ;
			solve(1 , tot , 1 , tot) ;
			int p = 0 ;
			for(int k = j ; k <= m ; k += i)
				dp[k] = f[++ p] ;
		}
		rep(j , 1 , m) cmax(dp[j] , dp[j - 1]) ;
	}
	rep(i , 1 , m) out << dp[i] << ' ' ;
	return 0;
  // code end.
}