『题解』Codeforces220B Little Elephant and Array

叁歲伎倆 2021-10-29 08:42 150阅读 0赞

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### 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; }

转载于:https://www.cnblogs.com/shenxiaohuang/p/11302408.html

[Link 1]: https://xiaohuang888.github.io/2019/08/05/%E3%80%8E%E9%A2%98%E8%A7%A3%E3%80%8FCodeforces220B%20Little%20Elephant%20and%20Array
[Codeforces]: http://codeforces.com/problemset/problem/220/B
[Luogu]: https://www.luogu.org/problem/CF220B