链接：

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 
     
      using namespace std;const double EPS = 1e-6;const int INF = 0x3f3f3f3f;const int mod = 1e9 + 7;const int maxn = 1e5 + 10;int n, m, x, y;int a[maxn];char q[maxn];struct Node{
   int mi, sum, add;} T[maxn<<2];void Build(int l, int r, int rt){    if(l == r){        T[rt].mi = a[l];        T[rt].sum = 0;        T[rt].add = 0;        return ;    }    int mid = (l + r) >> 1;    Build(l, mid, rt<<1);    Build(mid + 1, r, rt<<1|1);    T[rt].mi = min(T[rt<<1].mi, T[rt<<1|1].mi);    T[rt].sum = T[rt<<1].sum + T[rt<<1|1].sum;    T[rt].add = T[rt<<1].add + T[rt<<1|1].add;}void Pushdown(int rt, int l, int r){    if(!T[rt].add) return;    int mid = (l + r) >> 1;    T[rt<<1].mi += T[rt].add;    T[rt<<1|1].mi += T[rt].add;    T[rt<<1].add += T[rt].add;    T[rt<<1|1].add += T[rt].add;    T[rt].add = 0;}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 > 0) return;        if(l == r){            if(T[rt].mi == 0) T[rt].sum++, T[rt].mi = a[l];            return;        }        ok = false;    }    Pushdown(rt, l, r);    int mid = (l + r) >> 1;    if(L <= mid) Update(L, R, l, mid, rt<<1, ok);    if(R > mid) Update(L, R, mid + 1, r, rt<<1|1, ok);    T[rt].mi = min(T[rt<<1].mi, T[rt<<1|1].mi);    T[rt].sum = T[rt<<1].sum + T[rt<<1|1].sum;}int Query(int L, int R, int l, int r, int rt){    if(L <= l && r <= R) return T[rt].sum;    int ans = 0, mid = (l + r) >> 1;    if(L <= mid) ans += Query(L, R, l, mid, rt<<1);    if(R > mid) ans += Query(L, R, mid + 1, r, rt<<1|1);    return ans;}int main(){    while(scanf("%d%d", &n, &m) != EOF){        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);        Build(1, n, 1);        while(m--){            scanf("%s%d%d", q, &x, &y);            if(q[0] == 'a') Update(x, y, 1, n, 1, true);            else printf("%d\n", Query(x, y, 1, n, 1));        }    }    return 0;}