听说是线段树上套二分，可是不知道怎么写。

此份代码 $60$ pts。

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef __int128 i128;
const int N = 2e5 + 10;
int n,q;
ll s,b[N];
i128 w;
i128 a[N],sum[N << 2],lzy[N << 2];
void pushup(int o){
	sum[o] = sum[o << 1] + sum[o << 1 | 1];
}
void build(int o,int l,int r){
	if (l == r){
		sum[o] = a[l];
		return;
	}
	int mid = (l + r) >> 1;
	build(o << 1,l,mid);
	build(o << 1 | 1,mid + 1,r);
	pushup(o);
}
void pushdown(int o,int l,int r){
	int mid = (l + r) >> 1;
	sum[o << 1] += lzy[o] * (mid - l + 1);
	sum[o << 1 | 1] += lzy[o] * (r - mid);
	lzy[o << 1] += lzy[o];
	lzy[o << 1 | 1] += lzy[o];
	lzy[o] = 0;
}
void update(int o,int l,int r,int ql,int qr,i128 x){
	if (ql <= l && r <= qr){
		sum[o] += (r - l + 1) * x;
		lzy[o] += x;
		return;
	}
	int mid = (l + r) >> 1;
	pushdown(o,l,r);
	if (ql <= mid) update(o << 1,l,mid,ql,qr,x);
	if (qr > mid) update(o << 1 | 1,mid + 1,r,ql,qr,x);
	pushup(o);
}
i128 query(int o,int l,int r,int ql,int qr){
	if (ql <= l && r <= qr)
	    return sum[o];
	int mid = (l + r) >> 1;
	i128 ret = 0;
	pushdown(o,l,r);
	if (ql <= mid) ret += query(o << 1,l,mid,ql,qr);
	if (qr > mid) ret += query(o << 1 | 1,mid + 1,r,ql,qr);
	return ret;
}
int main(){
	freopen("wxyt4.in","r",stdin);
	freopen("wxyt4.out","w",stdout);
	ios::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin >> n >> q >> s;
	w = s;
	for (int i = 1 ; i <= n ; i++) cin >> b[i];
	for (int i = 1 ; i <= n ; i++) a[i] = b[i];
	build(1,1,n);
	for (int i = 1 ; i <= q ; i++){
		int l,r;
		ll d;
		i128 e;
		cin >> l >> r >> d;
		e = d;
		update(1,1,n,l,r,e);
		ll level = 1;
		int ans = 0;
		i128 blood = w;
		i128 all = query(1,1,n,1,n);
		while(1){
			if (blood <= all) break;
			blood -= all;
			all *= 2;
			ans += n;
			level *= 2;
		}
		int L = 1,R = n,tmid,p = 0;
		while(L <= R){
			tmid = (L + R) >> 1;
			if (query(1,1,n,1,tmid) * level < blood){
				p = tmid;
				L = tmid + 1;
			}else{
				R = tmid - 1;
			}
		}
		cout << ans + p << '\n';
	}
	return 0;
}