『题解』Codeforces220B Little Elephant and Array

Posted 2022-06-12 shenxiaohuang

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\(l_j\)和\(r_j (1 \le l_j \le r_j \le n)\)，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\n", ans[i]);
    return 0;