hdu6609(树状数组+二分)

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题意：给你一个长度为的数组和一个，求出对每一个最少删除 ~ 中多少个元素，使得剩余的 ~ 中的元素的和小于等于
代码：

#include <map>
#include <set>
#include <queue>
#include <string>
#include <math.h>
#include <vector>
#include <stdio.h>
#include <stdlib.h>
#include <iostream>
#include <string.h>
#include <algorithm>
using namespace std;
const int siz=2*100005;
map<int,int> mp;
int n,a[siz];
long long sum[siz],num[siz];
struct node{
    int u,id;
    friend bool operator<(node a,node b){
        if(a.u==b.u)
        return a.id<b.id;
        return a.u<b.u;
    }
};
vector<node> G;
int lowbit(int x){
    return x&(-x);
}
void update(int x,int y){
    while(x<=n){
        num[x]++;
        sum[x]+=y;
        x+=lowbit(x);
    }
}
long long query_sum(int x){
    long long ans=0;
    while(x>0){
        ans+=sum[x];
        x-=lowbit(x);
    }
    return ans;
}
long long query_num(int x){
    long long ans=0;
    while(x>0){
        ans+=num[x];
        x-=lowbit(x);
    }
    return ans;
}
int main(){
    int i,j,u,t,l,r,ans,mid;
    long long m,tmp;
    scanf("%d",&t);
    while(t--){                                 //sum按zhi值维护前缀和，num按个数维护前缀和
        scanf("%d%lld",&n,&m);
        memset(sum,0,sizeof(sum));
        memset(num,0,sizeof(num));
        G.clear();
        for(i=1;i<=n;i++){
            scanf("%d",&a[i]);
            G.push_back((node){a[i],i});
        }
        sort(G.begin(),G.end());
        for(i=0;i<G.size();i++)
        mp[G[i].id]=i+1;
        for(i=1;i<=n;i++){                      //每次二分出最长小于等于m-当前值的最长前缀的长度
            ans=0;                              //算出当前前缀中所含元素的个数
            l=1,r=n,tmp=m-a[i];
            while(l<=r){
                mid=(l+r)>>1;
                if(query_sum(mid)<=tmp){
                    ans=mid;
                    l=mid+1;
                }
                else
                r=mid-1;
            }
            printf("%d ",i-1-query_num(ans));
            update(mp[i],a[i]);
        }
        printf("\n");
    }
    return 0;
}
/*
2
5 10
10 10 10 10 10
*/