题目大意

题目链接Naive Operations

题目大意:

    区间加1(在a数组中)
   区间求ai/bi的和
   ai初值全部为0，bi给出，且为n的排列，多组数据(<=5),n,q<=1e5

axmorz

思路

因为是整除，所以一次加法可以对ans 没有影响

当ai是bi的倍数，对ans会有贡献

所以我们维护一个sum，初值为bi(只对于线段树的叶子节点有用)

当区间+1的时候

我们对sum-1

当sum=0的时候（倍数）

ans++，sum=bi

然后再维护一个区间最小值

当区间内的mi>0时

显然、对答案没有贡献

当区间内的mi<=0时

对答案一定有贡献

所以再在l到r内进行递归

对于g()函数的复杂度为

ans=n/1+n/2+n/3+n/4````+n/n=nlogn级别（看的别的题解护臂逼的）

代码

#include <iostream>
#include <cstring>
#include <cstdio>
#define ls rt<<1
#define rs rt<<1|1
using namespace std;
const int maxn = 1e5 + 7;
const int maxm = 4e5 + 7;
const int inf = 0x3f3f3f3f;

int n, m, b[maxn];
struct node {
    int l, r;
    int ans, sum, mi, lazy;
} e[maxm];

int read() {
    int x = 0, f = 1; char s = getchar();
    for (; s < '0' || s > '9'; s = getchar()) if (s == '-') f = -1;
    for (; s >= '0' && s <= '9'; s = getchar()) x = x * 10 + s - '0';
    return x * f;
}

void pushup(int rt) {
    e[rt].ans = e[ls].ans + e[rs].ans;
    e[rt].mi = min(e[ls].mi, e[rs].mi);
}
void pushdown(int rt) {
    if (e[rt].lazy) {
        e[ls].lazy += e[rt].lazy;
        e[rs].lazy += e[rt].lazy;
        e[ls].mi -= e[rt].lazy;
        e[rs].mi -= e[rt].lazy;

        e[rt].lazy = 0;
    }
}

void build(int l, int r, int rt) {
    e[rt].l = l, e[rt].r = r;
    if (l == r) {
        e[rt].sum = e[rt].mi = b[l];
        return;
    }
    int mid = (l + r) >> 1;
    build(l, mid, ls);
    build(mid + 1, r, rs);
    pushup(rt);
}

void g(int L, int R, int rt) {
    if (e[rt].mi > 0) {
        return ;
    } else {
        if (e[rt].l == e[rt].r) {
            e[rt].ans++;
            e[rt].mi = e[rt].sum = b[e[rt].l];
            return ;
        }
    }
    pushdown(rt);
    if (e[ls].mi <= 0) g(L, R, ls);
    if (e[rs].mi <= 0) g(L, R, rs);
    pushup(rt);
}

void add(int L, int R, int rt) {
    if (L <= e[rt].l && e[rt].r <= R) {
        e[rt].mi--;
        e[rt].lazy++;
        g(e[rt].l, e[rt].r, rt);
        //cout<<e[rt].l<<" "<<e[rt].r<<" "<<e[rt].mi<<"<M<<<<<\n";
        return;
    }
    pushdown(rt);
    int mid = (e[rt].l + e[rt].r) >> 1;
    if (L <= mid) add(L, R, ls);
    if (R > mid) add(L, R, rs);
    pushup(rt);
}

void debug() {
    printf("debug\n");
    printf("               %d\n", e[1].mi);
    printf("       %d               %d\n", e[2].mi, e[3].mi );
    printf("   %d       %d       %d       %d\n", e[4].mi, e[5].mi, e[6].mi, e[7].mi );
    printf(" %d   %d   %d   %d   %d   %d   %d   %d\n", e[8].mi,
     e[9].mi, e[10].mi, e[11].mi, e[12].mi, e[13].mi, e[14].mi, e[15].mi);
}

int query(int L, int R, int rt) {
    if (L <= e[rt].l && e[rt].r <= R) {
        return e[rt].ans;
    }
    pushdown(rt);
    int mid = (e[rt].l + e[rt].r) >> 1, ans = 0;
    if (L <= mid) ans += query(L, R, ls);
    if (R > mid) ans += query(L, R, rs);
    pushup(rt);
    return ans;
}

int main() {
    while(scanf("%d%d",&n,&m)!=EOF) {
        for (int i = 1; i <= n; ++i)
        b[i] = read();
        memset(e,0,sizeof(e));
        build(1, n, 1);
        for (int i = 1; i <= m; ++i) {
            char s[10];
            scanf("%s", s);
            int a = read(), b = read();
            if (s[0] == 'a') {
                add(a, b, 1);
            } else {
                printf("%d\n", query(a, b, 1));
            }
        }
    }
    return 0;
}