zoukankan      html  css  js  c++  java
  • 『题解』Codeforces220B Little Elephant and Array

    更好的阅读体验

    Portal

    Portal1: Codeforces

    Portal2: Luogu

    Description

    The Little Elephant loves playing with arrays. He has array (a), consisting of (n) positive integers, indexed from (1) to (n). Let's denote the number with index (i) as (a_i).

    Additionally the Little Elephant has (m) queries to the array, each query is characterised by a pair of integers (l_j) and (r_j (1 le l_j le r_j le n)). For each query (l_j, r_j) the Little Elephant has to count, how many numbers (x) exist, such that number (x) occurs exactly (x) times among numbers (a_{l_j}, a_{l_j + 1}, cdots , a_{r_j}).

    Help the Little Elephant to count the answers to all queries.

    Input

    The first line contains two space-separated integers (n) and (m (1 le n, m le 105)) - the size of array a and the number of queries to it. The next line contains (n) space-separated positive integers (a_1, a_2, cdots , a_n (1 le ai le 10^9)). Next (m) lines contain descriptions of queries, one per line. The (j)-th of these lines contains the description of the (j)-th query as two space-separated integers (l_j) and (r_j (1 le l_j le r_j le n)).

    Output

    In (m) lines print (m) integers - the answers to the queries. The (j)-th line should contain the answer to the (j)-th query.

    Sample Input1

    7 2
    3 1 2 2 3 3 7
    1 7
    3 4
    

    Sample Output

    3
    1
    

    Description in Chinese

    小象喜欢和数组玩。现在有一个数组(a),含有(n)个正整数,记第(i)个数为(a_i)

    现在有(m)个询问,每个询问包含两个正整数(l_j)(r_j (1 le l_j le r_j le n)),小象想知道在(a_{l_j}到)a_{r_j}(之中有多少个数)x(,其出现次数也为)x$。

    Solution

    我们先看题目,发现只有查询,没有修改,所以可以用普通的莫队解决。

    题目中的(a_i)的范围是(in [1, 10^9]),而数的总数的范围是(in [1, 10^5]),所以当这个数大于(10^5)了,这个数就不可能为所求的(x)忽略这个数后就不用进行离散化了。

    Code

    #include<iostream>
    #include<algorithm>
    #include<cstdio>
    #include<cstring>
    #include<cmath>
    
    using namespace std;
    
    const int MAXN = 100005;
    int n, m, nowans, a[MAXN], bl[MAXN], ans[MAXN], cnt[MAXN];
    struct node {
        int l, r, id;
        bool operator < (const node &x) const {//排序的规则
            return bl[l] == bl[x.l] ? r < x.r : bl[l] < bl[x.l];
        }
    } info[MAXN];
    inline void add(int x) {
        if (a[x] > n) return ;//如果数值超了n的范围就退出
        if (cnt[a[x]] == a[x]) nowans--;
        cnt[a[x]]++;
        if (cnt[a[x]] == a[x]) nowans++;
    }
    inline void dec(int x) {
        if (a[x] > n) return ;
        if (cnt[a[x]] == a[x]) nowans--;
        cnt[a[x]]--;
        if (cnt[a[x]] == a[x]) nowans++;
    }
    int main() {
        scanf("%d%d", &n, &m);
        int block = (int)sqrt(n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            bl[i] = i / block;
        }
        for (int i = 1; i <= m; i++) {
            scanf("%d%d", &info[i].l, &info[i].r);//莫队是离线算法,需要记录询问的左右端点
            info[i].id = i;//记录每个询问的编号
        }
        sort(info + 1, info + m + 1);
        memset(cnt, 0, sizeof(cnt));
        int    l = 1, r = 0;
        for (int i = 1; i <= m; i++) {//莫队
            while (l < info[i].l) dec(l++);
            while (l > info[i].l) add(--l);
            while (r < info[i].r) add(++r);
            while (r > info[i].r) dec(r--);
            ans[info[i].id] = nowans;
        }
        for (int i = 1; i <= m; i++)
            printf("%d
    ", ans[i]);
        return 0;
    }
    
  • 相关阅读:
    Helvetic Coding Contest 2016 online mirror D1
    Helvetic Coding Contest 2016 online mirror C1
    Helvetic Coding Contest 2016 online mirror A1
    Educational Codeforces Round 13 C
    Educational Codeforces Round 13 B
    Educational Codeforces Round 13 A
    2016计蒜之道初赛第四场A
    帆软出品: 7点搞定制药企业数据分析系统开发需求
    制药企业BI系统方案整体设计分享
    Ubuntu ROS Arduino Gazebo学习镜像iso说明(indigo版)
  • 原文地址:https://www.cnblogs.com/shenxiaohuang/p/11302408.html
Copyright © 2011-2022 走看看