【HDOJ6609】Find the answer（线段树）

题意：给定一个n个正整数的数列，第i项为w[i]，对于每个i，你要从[1,i-1]中选择一些变成0，使得变化后[1,i]的总和小于m，每次询问最少要变几个

n<=2e5，m<=1e9，1<=w[i]<=m

思路：显然每次贪心删最大的，直接开权值线段树，每次询问就在直接树上二分

开始交了几发TLE+MLE，是没有离散化的锅，把w[i]离散化就行

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
#define N  210000
#define M  1100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=998244353,inv2=(MOD+1)/2;
      double eps=1e-6;
      ll INF=1e14;

struct node
{
    int num;
    ll sum;
}t[N<<2];

struct arr
{
    int x,id;
}a[N];

bool cmp(arr a,arr b)
{
    return a.x<b.x;
}

int b[N],c[N],q[N];

int cnt,root,n,m;

int read()
{
   int v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

void build(int l,int r,int p)
{
    t[p].num=t[p].sum=0;
    if(l==r) return;
    int mid=(l+r)>>1;
    build(l,mid,ls);
    build(mid+1,r,rs);
}

int query(int l,int r,ll s,int p)
{
    if(p==0) return 0;
    if(l==r)
    {
        if(s<=m) return 0;
        ll t=(s-m)/c[l];
        if((s-m)%c[l]) t++;
        return t;
    }
    int mid=(l+r)>>1;
    ll tmp=t[rs].sum;
    if(s-tmp<=m) return query(mid+1,r,s,rs);
     else
     {
         return t[rs].num+query(l,mid,s-tmp,ls);
     }
}

void update(int l,int r,int x,int p)
{
    t[p].num++;
    t[p].sum+=c[x];
    if(l==r) return;
    int mid=(l+r)>>1;
    if(x<=mid) update(l,mid,x,ls);
     else update(mid+1,r,x,rs);
}

int main()
{
    //freopen("1.in","r",stdin);
    //freopen("1.out","w",stdout);
    int cas=read();
    cnt=0;
    while(cas--)
    {
        n=read(),m=read();
        rep(i,1,n)
        {
            a[i].x=read();
            a[i].id=i;
        }
        rep(i,1,n) q[i]=a[i].x;
        sort(a+1,a+n+1,cmp);
        b[a[1].id]=1; c[1]=a[1].x;
        int t=1;
        rep(i,2,n)
         if(a[i].x==a[i-1].x) b[a[i].id]=t;
          else
          {
              t++;
              b[a[i].id]=t;
              c[t]=a[i].x;
          }
        build(1,t,1);
        ll s=0;
        rep(i,1,n)
        {
            s+=q[i];
            printf("%d ",query(1,t,s,1));
            update(1,t,b[i],1);
        }
        //printf("cnt=%d
",cnt);
        printf("
");
    }
    return 0;
}