Luo Gu [P1972] [HH] BZOJ1878 necklace [Mo] team

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Subject to the effect:

Topic links:
Luo Gu: https://www.luogu.org/problemnew/show/P1972
BZOJ: https://www.lydsy.com/JudgeOnline/problem.php?id=1878

A given sequence of query $[l_i,r_i]$ How many different numbers.

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Ideas:

Mo Bare team title. In fact, this question is a weaker version (without modification)
but this question Alcamo right. open $Ofast,O3$ will MLE.
Do not talk about the team method, I believe we will. Here to talk about how the team card with Mo over this question.

1. Input Output Optimization
2. Block length $T=1250$
3. Merge statement.
   This can greatly improve the efficiency of the program. In addition to the code are okay wind ugly accident

So far the program can be snapped into the $1200ms$ . Mo is the most original team $2000+ms$ is.
Really do not know how to cut. So Mo had looked at the card team code.
They find that sort of no avail $cmp$ , but staged this

bool operator < (const Ask &a,const Ask &b){
	return pos[a.l]^pos[b.l] ? a.l<b.l : pos[a.l]&1 ? a.r<b.r:a.r>b.r;
}

Seen from the conditional statement, this apparently is in place $cmp$ a.
And can greatly speed up. I do not know the specific reasons $moment$ .

4. Plus the metaphysics optimization.
5. Luo open valley comes $O2$ .

This can be very stable over this question out of.

We can see the slowest point $600ms$ inside, considered a fast.

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Code:

// luogu-judger-enable-o2
#include <cmath>
#include <ctime>
#include <cstdio>
#include <string>
#include <algorithm>
#define reg register
using namespace std;

const int N=500010,M=1000010;
int n,m,l,r,sum,T,a[N],cnt[M],pos[N],ans[N];

struct Ask
{
    int l,r,id;
}ask[N];

bool operator < (const Ask &a,const Ask &b)
{
    return pos[a.l]^pos[b.l] ? a.l<b.l : pos[a.l]&1 ? a.r<b.r:a.r>b.r;
}

inline int read()
{
    int d=0;
    char ch=getchar();
    while (!isdigit(ch)) ch=getchar();
    while (isdigit(ch))
        d=(d<<3)+(d<<1)+ch-48,ch=getchar();
    return d;
}

inline void write(int x)
{
    if (x>9) write(x/10);
    putchar(x%10+48);
}

int main()
{
    n=read();
    T=1250;
    for (reg int i=1;i<=n;++i)
    {
        a[i]=read();
        pos[i]=(i-1)/T+1;
    }
    m=read();
    for (reg int i=1;i<=m;++i)
    {
        ask[i].l=read(); ask[i].r=read();
        ask[i].id=i;
    }
    sort(ask+1,ask+1+m);
    l=1;
    for (reg int i=1;i<=m;++i)
    {
        while(l>ask[i].l) sum+=(++cnt[a[--l]]==1);
        while(l<ask[i].l) sum-=(--cnt[a[l++]]==0);
        while(r>ask[i].r) sum-=(--cnt[a[r--]]==0);
        while(r<ask[i].r) sum+=(++cnt[a[++r]]==1);
        ans[ask[i].id]=sum;
    }
    for (reg int i=1;i<=m;++i)
        write(ans[i]),putchar(10);
    return 0;
}