[bzoj3671][uoj6][Noi2014]随机数生成器【常数优化】【贪心】

【题目链接】
　　https://www.lydsy.com/JudgeOnline/problem.php?id=3671
　　http://uoj.ac/problem/6
【题解】
　　首先求出X的数列，这是这道题的难点（？），要注意尽量少取模。
　　接下来就简单了，从小到大枚举每个数，用贪心的策略选取，并把这个数的左下和右上方的所有数标记为不可选取，注意一旦标记到的数已经不可取了，那么直接退出循环，这样才能保证复杂度。
　　时间复杂度$O(N*M)$
【代码】

/* - - - - - - - - - - - - - - -
    User :      VanishD
    problem :
    Points :    
- - - - - - - - - - - - - - - */
/* - - - - - - - - - - - - - - -
    User :      VanishD
    problem :   [bzoj3671][uoj6][noi2014] 
    Points :    read?
- - - - - - - - - - - - - - - */
# include <bits/stdc++.h>
# define    ll      long long
# define    inf     0x3f3f3f3f
# define    N       5010
using namespace std;
int read(){
    int tmp = 0, fh = 1; char ch = getchar();
    while (ch < '0' || ch > '9'){ if (ch == '-') fh = -1; ch = getchar(); }
    while (ch >= '0' && ch <= '9'){ tmp = tmp * 10 + ch - '0'; ch = getchar(); }
    return tmp * fh;
}
int mp[N][N], n, m, q, ans[N * 2], ansnum, p[N * N], canx[N], cany[N];
int main(){
//  freopen(".in", "r", stdin);
//  freopen("A.out", "w", stdout);
    ll x = read(), a = read(), b = read(), c = read(), d = read();
    n = read(), m = read(), q = read();
    for (int i = 1, nowx = 1, nowy = 1; i <= n * m; i++){
        x = (a * x * x  + b * x + c) % d;
        mp[nowx][nowy] = i;
        int y = x % i + 1;
        int tmpx = (y - 1) / m + 1, tmpy = (y - 1) % m + 1;
        swap(mp[tmpx][tmpy], mp[nowx][nowy]);
        nowy++;
        if (nowy > m) nowx++, nowy = 1;
    }
    for (int i = 1; i <= q; i++){
        int tmp1 = read(), tmp2 = read(),
            tx1 = (tmp1 - 1) / m + 1, tx2 = (tmp2 - 1) / m + 1,
            ty1 = (tmp1 - 1) % m + 1, ty2 = (tmp2 - 1) % m + 1;
        swap(mp[tx1][ty1], mp[tx2][ty2]);
    }
//  for (int i = 1; i <= n; i++)
//      for (int j = 1; j <= m; j++)
//          printf("%d%c", mp[i][j], (j == m) ? '\n' : ' ');
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
            p[mp[i][j]] = (i - 1) * m + j;
    for (int i = 1; i <= n; i++)
        canx[i] = 0, cany[i] = m + 1;
    for (int i = 1; i <= n * m; i++){
        int x = (p[i] - 1) / m + 1, y = (p[i] - 1) % m + 1;
        if (canx[x] < y && cany[x] > y){
            ans[++ansnum] = i;
            for (int tx = x + 1; tx <= n; tx++){
                if (canx[tx] >= y - 1) break;
                canx[tx] = y - 1;
            }
            for (int tx = x - 1; tx >= 1; tx--){
                if (cany[tx] <= y + 1) break;
                cany[tx] = y + 1;
            }
        }
    }
    for (int i = 1; i <= ansnum; i++)
        printf("%d%c", ans[i], (i == ansnum) ? '\n' : ' ');
    return 0;
}