Ultraviolet A 12003 array Transformer (block array)

Page 1 of the White Paper.

Always messing around on the original array. In fact, we need to use another array record block.

The original array cannot be changed.

Pay attention to the relationship between the original array and the block array, and carefully process the boundary. It is not difficult.

#include <cstdio>#include <iostream>#include <algorithm>#include <cstring>#define maxn 300005#define SIZE 600using namespace std;int a[maxn];int block[maxn];int b[maxn];void work(int p,int x){    int old=b[p];    b[p]=x;    int BLOCK=p/SIZE;    int pos=p;    for(int i=BLOCK*SIZE;i<BLOCK*SIZE+SIZE;i++)    {        if(a[i]==old)        {            pos=i;            a[i]=x;            break;        }    }    while(block[pos+1]==block[p] && a[pos+1]<a[pos])    {        swap(a[pos+1],a[pos]);        pos++;    }    while(pos-1>=0 && block[pos-1]==block[p] && a[pos-1]>a[pos])    {        swap(a[pos-1],a[pos]);        pos--;    }}int main(){    int n,m,u;    while(scanf("%d%d%d",&n,&m,&u)!=EOF)    {        memset(a,0x3f,sizeof a);        memset(block,0x3f,sizeof block);        for(int i=0;i<n;i++)        {            scanf("%d",&a[i]);            b[i]=a[i];            block[i]=i/SIZE;        }        for(int i=0;i<(n-1)/SIZE;i++)        {            sort(a+i*SIZE,a+(i+1)*SIZE);        }        sort(a+((n-1)/SIZE*SIZE),a+n);        while(m--)        {            int l,r,v,p;            scanf("%d%d%d%d",&l,&r,&v,&p);            l--,r--,p--;            int ans=0;            if(block[l]==block[r])            {                for(int i=l;i<=r;i++)                if(b[i]<v)ans++;            }            else            {                for(int i=l;block[i]==block[l];i++)                {                    if(b[i]<v)ans++;                }                for(int i=r;block[i]==block[r];i--)                if(b[i]<v)ans++;                for(int i=SIZE*(l/SIZE+1);i<r/SIZE*SIZE;i+=SIZE)                {                    ans+=lower_bound(a+i,a+i+SIZE,v)-(a+i);                }            }            work(p,(long long)u*ans/(r-l+1));        }        for(int i=0;i<n;i++)        printf("%d\n",b[i]);    }    return 0;}/*10 3 510 9 8 7 6 5 4 3 2 12 5 9 42 9 6 54 10 5 6*/