BZOJ 4320: ShangHai2006 Homework

考虑 分块

令 $unit = sqrt(300000)$

对于 小于$x <= unit 的 直接暴力O(unit)维护，O(1) 查询$

对于 大于$x > unit 的 考虑枚举 y = k cdot x, 再查询 [y, 2y]之间的出现过的最小值，此时的答案为 最小值 - y$

$暴力枚举上去 发现是O(unit)的查询，再考虑维护$

$可以对权值分块进行维护，块内暴力维护，块间作标记$

总的复杂度为$O(nsqrt{300000})$

#include <bits/stdc++.h>
using namespace std;

#define N 100010
#define M 300010
#define unit 450
int n = 300000, q, f[unit << 1], g[M], h[unit << 1];

void add(int x) 
{
    for (int i = 1; i <= unit; ++i) f[i] = min(f[i], x % i); 
    int st = ((x - 1) / unit) * unit + 1, ed = (x - 1) / unit;
    for (int i = st; i <= x; ++i) g[i] = min(g[i], x);
    for (int i = 0; i < ed; ++i) h[i] = min(h[i], x);
}

int ask(int x){ return min(g[x], h[(x - 1) / unit]); }

int query(int x)
{ 
    if (x <= unit) return f[x];  
    int res = M; 
    for (int i = 0, j = x; i <= n; j += x)
    {
        int t = ask(max(i, 1)); 
        res = min(res, t - i); 
        i = j; 
    }
    return res;
}

void init()
{
    memset(f, 0x3f, sizeof f);
    memset(g, 0x3f, sizeof g);
    memset(h, 0x3f, sizeof h);
}

void Run()
{
    while (scanf("%d", &q) != EOF)
    {
        init();
        char op[10]; int x;
        for (int qq = 1; qq <= q; ++qq)
        {
            scanf("%s%d", op, &x);
            if (op[0] == 'A') add(x);
            else printf("%d
", query(x));
        }
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif 

    Run();
    return 0;
}