hdu 6609 区间条件前缀和 + 二分

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("
");
	}
}