QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#24673#1877. Matryoshka Dollszzy0922WA 20ms17248kbC++2.7kb2022-04-02 08:20:562022-04-30 06:26:34

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-04-30 06:26:34]
  • 评测
  • 测评结果:WA
  • 用时:20ms
  • 内存:17248kb
  • [2022-04-02 08:20:56]
  • 提交

answer

#include <bits/stdc++.h>

using namespace std;

#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1<<21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 21],*p1 = buf,*p2 = buf;
inline int read()
{
    char c = getchar();
    int x = 0;
    bool f = 0;
    for(;!isdigit(c); c = getchar()) f ^= !(c ^ 45);
    for(;isdigit(c); c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
    if (f) x = -x; 
    return x;
}

int n, m, a[500005], b[500005], pos[500005];
long long anslist[500005];
int len;
struct query{
    int l, r, idx;
    bool operator<(query a) {
        if (l / len == a.l / len) return r > a.r;
        else return l / len < a.l / len; 
    }
}q[500005];

struct node {
    int val, next, prev;
}lis[500005];
#define head lis[0]
#define tail lis[n + 1]
long long ans, tmp, now;

void del(int p, long long& x) {
    if (lis[p].prev == 0) x -= abs(p - lis[p].next);
    else if (lis[p].next == n + 1) x -= abs(p - lis[p].prev);
    else if (lis[p].prev == 0 && lis[p].next == n + 1) x = 0;
    else x = x - abs(p - lis[p].prev) - abs(p - lis[p].next) + abs(lis[p].next - lis[p].prev);
    lis[lis[p].prev].next = lis[p].next;
    lis[lis[p].next].prev = lis[p].prev;
}

void rcv(int p) {
    lis[lis[p].prev].next = p;
    lis[lis[p].next].prev = p;
}

int main() {
    scanf("%d%d", &n, &m);
    len = sqrt(n);
    for (int i = 1; i <= n; i++)
        scanf("%d", &a[i]);
    for (int i = 1; i <= n; i++)
        lis[i].val = a[i];
    for (int i = 1; i <= n; i++)
        pos[a[i]] = i;
    memcpy(b, a, sizeof a);
    sort(b + 1, b + n + 1);
    for (int i = 1; i < n; i++) {
        ans += abs(pos[b[i]] - pos[b[i + 1]]);
    }
    head.next = pos[b[1]];
    lis[pos[b[1]]].prev = 0;
    for (int i = 1; i < n; i++) {
        lis[pos[b[i]]].next = pos[b[i + 1]];
        lis[pos[b[i + 1]]].prev = pos[b[i]];
    }
    tail.prev = pos[b[n]];
    lis[pos[b[n]]].next = n + 1;
    for (int i = 1; i <= m; i++)
        scanf("%d%d", &q[i].l, &q[i].r);
    for (int i = 1; i <= m; i++)
        q[i].idx = i;
    sort (q + 1, q + m + 1);
    now = ans;
    //cout << ans;
    for (int i = 1, l = 1, r = n; i <= m; i++) {
        // if (q[i].l / len != q[i - 1].l /len) {
        //     while (l < (q[i].l - 1) / len * len + 1) del(l++, now);
        //     r = n;
        //     tmp = ans = now;
        // }
        l = 1, r = n;
        ans = now;
        while (r > q[i].r) del(r--, ans);
        tmp = ans;
        while (l < q[i].l) del(l++, tmp);
        anslist[q[i].idx] = tmp;
        while (l > (q[i].l - 1) / len * len + 1) rcv(l--);
    }
    for (int i = 1; i <= m; i++) printf("%lld\n", anslist[i]);
    return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 2ms
memory: 13956kb

input:

5 5
1 5 2 4 3
1 5
1 4
1 3
1 2
1 1

output:

7
5
3
1
0

result:

ok 5 number(s): "7 5 3 1 0"

Test #2:

score: 0
Accepted
time: 0ms
memory: 14016kb

input:

1 1
1
1 1

output:

0

result:

ok 1 number(s): "0"

Test #3:

score: 0
Accepted
time: 20ms
memory: 17248kb

input:

100000 1
2 4 6 8 10 12 14 16 18 20 22 24 26 28 30 32 34 36 38 40 42 44 46 48 50 52 54 56 58 60 62 64 66 68 70 72 74 76 78 80 82 84 86 88 90 92 94 96 98 100 102 104 106 108 110 112 114 116 118 120 122 124 126 128 130 132 134 136 138 140 142 144 146 148 150 152 154 156 158 160 162 164 166 168 170 172 ...

output:

4999950000

result:

ok 1 number(s): "4999950000"

Test #4:

score: 0
Accepted
time: 2ms
memory: 16080kb

input:

20 1
12 8 13 10 18 14 1 19 5 16 15 9 17 20 6 2 11 4 3 7
9 18

output:

36

result:

ok 1 number(s): "36"

Test #5:

score: -100
Wrong Answer
time: 0ms
memory: 14016kb

input:

20 10
5 16 11 7 19 8 12 13 17 18 6 1 14 3 4 2 15 20 10 9
7 11
7 13
7 17
11 15
1 7
4 6
1 5
6 14
3 5
9 9

output:

25
35
72
30
20
3
8
41
3
3

result:

wrong answer 1st numbers differ - expected: '7', found: '25'