HDU-6315：Naive Operations（线段树+思维）

链接：HDU-6315：Naive Operations

题意：

In a galaxy far, far away, there are two integer sequence a and b of length n.
b is a static permutation of 1 to n. Initially a is filled with zeroes.
There are two kind of operations:
1. add l r: add one for al,al+1...ar
2. query l r: query ∑ri=l⌊ai/bi⌋
题解：每次更新时，直接将 b 数组的对应区间都减一，如果发现有零，就将对应位置的答案加一并重置回原b数组对应的数。

#include <bits/stdc++.h>
using namespace std;

const double EPS = 1e-;
const int INF = 0x3f3f3f3f;
const int mod = 1e9 + ;
const int maxn = 1e5 + ;
int n, m, x, y;
int a[maxn];
char q[maxn];
struct Node{int mi, sum, add;} T[maxn<<];

void Build(int l, int r, int rt)
{
    if(l == r){
        T[rt].mi = a[l];
        T[rt].sum = ;
        T[rt].add = ;
        return ;
    }

    int mid = (l + r) >> ;

    Build(l, mid, rt<<);
    Build(mid + , r, rt<<|);

    T[rt].mi = min(T[rt<<].mi, T[rt<<|].mi);
    T[rt].sum = T[rt<<].sum + T[rt<<|].sum;
    T[rt].add = T[rt<<].add + T[rt<<|].add;
}

void Pushdown(int rt, int l, int r)
{
    if(!T[rt].add) return;

    int mid = (l + r) >> ;

    T[rt<<].mi += T[rt].add;
    T[rt<<|].mi += T[rt].add;

    T[rt<<].add += T[rt].add;
    T[rt<<|].add += T[rt].add;

    T[rt].add = ;
}

void Update(int L, int R, int l, int r, int rt, bool ok)
{
    if(L <= l && r <= R){
        if(ok){
            T[rt].add--;
            T[rt].mi--;
        }
        if(T[rt].mi > ) return;
        if(l == r){
            if(T[rt].mi == ) T[rt].sum++, T[rt].mi = a[l];
            return;
        }
        ok = false;
    }
    Pushdown(rt, l, r);

    int mid = (l + r) >> ;

    if(L <= mid) Update(L, R, l, mid, rt<<, ok);
    if(R > mid) Update(L, R, mid + , r, rt<<|, ok);

    T[rt].mi = min(T[rt<<].mi, T[rt<<|].mi);
    T[rt].sum = T[rt<<].sum + T[rt<<|].sum;
}

int Query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R) return T[rt].sum;

    int ans = , mid = (l + r) >> ;

    if(L <= mid) ans += Query(L, R, l, mid, rt<<);
    if(R > mid) ans += Query(L, R, mid + , r, rt<<|);

    return ans;
}

int main()
{
    while(scanf("%d%d", &n, &m) != EOF){
        for(int i = ; i <= n; i++) scanf("%d", &a[i]);

        Build(, n, );

        while(m--){
            scanf("%s%d%d", q, &x, &y);
            if(q[] == 'a') Update(x, y, , n, , true);
            else printf("%d\n", Query(x, y, , n, ));
        }
    }

    return ;
}