Codeforces 220B（Little Elephant and Array）

131 阅读 0 评论 87 点赞

我是靠谱客的博主虚幻电话，这篇文章主要介绍Codeforces 220B（Little Elephant and Array），现在分享给大家，希望可以做个参考。

题意：给出一个长度为n的序列，然后给出m个区间询问，求区间内数字出现次数等于该数字的个数；

思路：这题有很多种解法，之前用一种nsqrt(n)的暴力解法写过，这次用莫队算法(分块)再写一次，复杂度是(n+q)sqrt(n)；

复制代码

#include <cmath>
#include <vector>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 100010;
int ans[N], num[N], L, R, cnt, M[N];
struct query {
    int l, r, idx, block;
    bool operator < (const query x) const {
        if(x.block == block)
            return r < x.r;
        return block < x.block;
    }
}q[N];

void add(int x) {
    if(x > 100000) return ;
    M[x]++;
    int occur = M[x];
    if(occur == x) cnt++;
    else if(occur == x + 1) cnt--;
}

void del(int x) {
    if(x > 100000) return ;
    M[x]--;
    int occur = M[x];
    if(occur == x) cnt++;
    else if(occur == x - 1) cnt--;
}

int main() {
    int n, nq;
    cin >> n >> nq;
    int BLOCK_SIZE = (int)sqrt(n);
    for(int i = 1; i <= n; i++)
        scanf("%d", &num[i]);
    for(int i = 1; i <= nq; i++) {
        scanf("%d%d", &q[i].l, &q[i].r);
        q[i].idx = i, q[i].block = q[i].l / BLOCK_SIZE;
    }

sort(q + 1, q + nq + 1);
    for(int i = q[1].l;i <= q[1].r;i++)
        add(num[i]);
    ans[q[1].idx] = cnt;
    L = q[1].l, R = q[1].r;
    for(int i = 2; i <= nq; i++) {
        for(int j = q[i].l; j < L; j++)
            add(num[j]);
        for(int j = L; j < q[i].l; j++)
            del(num[j]);
        for(int j = R + 1; j <= q[i].r; j++)
            add(num[j]);
        for(int j = q[i].r + 1; j <= R; j++)
            del(num[j]);
        L = q[i].l, R = q[i].r;
        ans[q[i].idx] = cnt;
    }
    for(int i = 1; i <= nq; i++)
        printf("%dn", ans[i]);
    return 0;
}

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
#include <cmath>
#include <vector>
#include <cstdio>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 100010;
int ans[N], num[N], L, R, cnt, M[N];
struct query {
    int l, r, idx, block;
    bool operator < (const query x) const {
        if(x.block == block)
            return r < x.r;
        return block < x.block;
    }
}q[N];

void add(int x) {
    if(x > 100000) return ;
    M[x]++;
    int occur = M[x];
    if(occur == x) cnt++;
    else if(occur == x + 1) cnt--;
}

void del(int x) {
    if(x > 100000) return ;
    M[x]--;
    int occur = M[x];
    if(occur == x) cnt++;
    else if(occur == x - 1) cnt--;
}

int main() {
    int n, nq;
    cin >> n >> nq;
    int BLOCK_SIZE = (int)sqrt(n);
    for(int i = 1; i <= n; i++)
        scanf("%d", &num[i]);
    for(int i = 1; i <= nq; i++) {
        scanf("%d%d", &q[i].l, &q[i].r);
        q[i].idx = i, q[i].block = q[i].l / BLOCK_SIZE;
    }

    sort(q + 1, q + nq + 1);
    for(int i = q[1].l;i <= q[1].r;i++)
        add(num[i]);
    ans[q[1].idx] = cnt;
    L = q[1].l, R = q[1].r;
    for(int i = 2; i <= nq; i++) {
        for(int j = q[i].l; j < L; j++)
            add(num[j]);
        for(int j = L; j < q[i].l; j++)
            del(num[j]);
        for(int j = R + 1; j <= q[i].r; j++)
            add(num[j]);
        for(int j = q[i].r + 1; j <= R; j++)
            del(num[j]);
        L = q[i].l, R = q[i].r;
        ans[q[i].idx] = cnt;
    }
    for(int i = 1; i <= nq; i++)
        printf("%dn", ans[i]);
    return 0;
}