4320: ShangHai2006 Homework

Posted 2021-02-04 mjtcn

4320: ShangHai2006 Homework

链接

分析：

　　分块。对权值模数进行分块，模数小于$sqrt V$的（$V$为权值上界），暴力处理。

　　模数大于$sqrt V$的，设模数是k，枚举k的倍数，然后查询大于[k,2k]之间的最小的数x，这个区间的mod k最小的数就是x-k。k的倍数共有$sqrt V$个，每次查询，再对权值进行分块，并维护后缀最小值，做到$O(1)$查询。复杂度$O(n sqrt V)$

代码：

#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
using namespace std;
typedef long long LL;

inline int read() {
    int x=0,f=1;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch==‘-‘)f=-1;
    for(;isdigit(ch);ch=getchar())x=x*10+ch-‘0‘;return x*f;
}

const int N = 300000, M = 550, INF = 1e9;
int bel[N + 5], tag[N + 5], f[N + 5], pt[N + 5];

void add(int x) {
    for (int i = 1; i <= M; ++i) f[i] = min(f[i], x % i);
    tag[bel[x]] = min(tag[bel[x]], x);
    pt[x] = min(pt[x], x);
    for (int i = x - 1, lim = (bel[x] - 1) * M; i >= lim; --i) pt[i] = min(pt[i], pt[i + 1]);
    for (int i = bel[x] - 1; i; --i) tag[i] = min(tag[i], tag[i + 1]);
}
int query(int x) {
    return min(pt[x], tag[bel[x] + 1]);
}
int Ask(int x) {
    if (x <= M) return f[x];
    int ans = query(1) % x;
    for (int i = x; i <= N; i += x) {
        int t = query(i) - i;
        if (t < x) ans = min(ans, t);
    }
    return ans;
}
int main() {
    memset(f, 0x3f, sizeof(f));
    memset(tag, 0x3f, sizeof(tag));
    memset(pt, 0x3f, sizeof(pt));
    int n = read();
    for (int i = 1; i <= N; ++i) bel[i] = (i - 1) / M + 1;
    char opt[10];
    for (int i = 1; i <= n; ++i) {
        scanf("%s", opt); int x = read();
        if (opt[0] == ‘A‘) add(x);
        else printf("%d
", Ask(x));
    }
    return 0;
}