hdu 6609 区间条件前缀和 + 二分

Posted 2022-05-26 tea-egg

题目传送门//res tp hdu

目的

在尾部逐步插入n个元素，求插入第i个元素时,[1,i)内删去多少个元素，可使前缀和[1,i]不大于m

多测Q [1,15]
n [1,2e5]
m [1,1e9]
每个元素W_i [1,m] (i∈[1,n]);

数据结构

树状数组

分析

维护两个树状数组，分别储存前缀和前缀中元素数量。先将元素全部读入，之后进行排列，同时记录好每个元素排列之后的下标。按下标向树状数组插入元素。之后二分枚举即可
时间复杂度O(Qnlognlogn)

#include<iostream>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int L = 200010;
ll val[L],sor[L];
struct E
    ll v;
    int pos;
sorted[L];
int Q,POS[L];
ll n,m;
ll BIT[L],bit[L];
int Len;
int lowbit(int x)return x&-x;
bool cmp(E a,E b)return a.v<b.v;
void change(int x,int y)
    sor[x] = y;
    for(int i = x;i<=L;i+=lowbit(i))
        BIT[i] += y;bit[i]++;
    

ll query(int k)//对前缀和的询问
    ll ans = 0;
    for(int i = k;i > 0; i -= lowbit(i))
        ans += BIT[i];
    return ans;

int query1(int k)//对前缀内元素个数的询问
    int ans = 0;
    for(int i = k;i>0;i-=lowbit(i))
        ans += bit[i];
    return ans;

int getans(int x,int M) //前x个元素的小序前缀之和不超过M
    int lo = 1,hi = n+1;
    int mi;
    ll sum;
    while(lo < hi)
        mi = (lo + hi)>>1;
        sum = query(mi);
        if(M <sum)  hi = mi;
        else lo = mi + 1;
    
    --lo;
    return x-query1(lo);


int main()
    scanf(" %d",&Q);
    while(Q--)
        scanf(" %lld %lld",&n,&m);
        for(int i = 1;i<=n;++i) BIT[i] = bit[i] = 0;
        for(int i = 1;i<=n;++i) sor[i] = 0;
        for(int i = 1;i<=n;++i) 
            scanf(" %lld",&val[i]);
            sorted[i].v = val[i];
            sorted[i].pos = i;
        
        sort(sorted+1,sorted+1+n,cmp);
        for(int i = 1;i<=n;++i) POS[sorted[i].pos] = i;
        int ans;
        for(int i = 1;i<=n;++i)
            ans = getans(i-1,m-val[i]);
            change(POS[i],val[i]);
            printf("%d ",ans);
        
        printf("\n");