Naive Operations HDU6315 （杭电多校2G）

让ci = ai / bi, 求sum(ci)的值，因为每次 ai 都是加一的，那么我可以用一颗线段树来维护每个 i 位置的 ai 距离达到 bi 还需要的数的最小值，更新是每次都减一，如果我某一个区间的最小值等于 0， 这就说明我这时候的ai已经满足了ai/bi==1的情况，那么对应的ci的位置就应该加一，所以我每次发现一个区间的最小值是0的话，我就 dfs 去找到为0的地方，把这里重新修改成bi，然后这个位置的答案+1就可以了

#include<map>
#include<set>
#include<ctime>
#include<cmath>
#include<stack>
#include<queue>
#include<string>
#include<vector>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<iostream>
#include<algorithm>
#define first fi
#define second se
#define lowbit(x) (x & (-x))

typedef unsigned long long int ull;
typedef long long int ll;
const double pi = 4.0*atan(1.0);
const int inf = 0x3f3f3f3f;
const int maxn = ;
const int maxm = ;
const int mod = ;
using namespace std;

int n, m, tol, T;
int a[maxn];
int b[maxn];
int c[maxn << ];
int sum[maxn << ];
int lazy[maxn << ];

void init() {
    memset(a, , sizeof a);
    memset(b, , sizeof b);
    memset(c, , sizeof c);
    memset(sum, inf, sizeof sum);
    memset(lazy, , sizeof lazy);
}

void pushup(int root) {
    sum[root] = min(sum[root << ], sum[root <<  | ]);
    c[root] = c[root << ] + c[root <<  | ];
}

void pushdown(int root) {
    if(lazy[root] != ) {
        lazy[root << ] += lazy[root];
        lazy[root <<  | ] += lazy[root];
        sum[root << ] -= lazy[root];
        sum[root <<  | ] -= lazy[root];
        lazy[root] = ;
    }
    return ;
}

void build(int left, int right, int root) {
    if(left == right) {
        sum[root] = b[left];
        return ;
    }
    int mid = (left + right) >> ;
    build(left, mid, root << );
    build(mid+, right, root <<  | );
    pushup(root);
}

void dfs(int left, int right, int root) {
    if(left == right) {
        if(sum[root] == ) {
            sum[root] = b[left];
            c[root]++;
        }
        return ;
    }
    pushdown(root);
    int mid = (left + right) >> ;
    if(sum[root << ] == )
        dfs(left, mid, root << );
    if(sum[root <<  | ] == )
        dfs(mid+, right, root <<  | );
    pushup(root);
}

void update(int left, int right, int prel, int prer, int root) {
    if(prel <= left && right <= prer) {
        lazy[root]++;
        sum[root]--;
        while(sum[root] == )    dfs(left, right, root);
        return ;
    }
    pushdown(root);
    int mid = (left + right) >> ;
    if(prel <= mid)    update(left, mid, prel, prer, root << );
    if(prer > mid)    update(mid+, right, prel, prer, root <<  | );
    pushup(root);
}

int query(int left, int right, int prel, int prer, int root) {
    if(prel <= left && right <= prer) return c[root];
    int mid = (left + right) >> ;
    int ans = ;
    if(prel <= mid)    ans += query(left, mid, prel, prer, root << );
    if(prer > mid)    ans += query(mid+, right, prel, prer, root <<  | );
    return ans;
}

int main() {
    while(~scanf("%d%d", &n, &m)) {
        init();
        for(int i=; i<=n; i++)    scanf("%d", &b[i]);
        build(, n, );
        char str[];
        while(m--) {
            int l, r;
            scanf("%s%d%d", str, &l, &r);
            if(str[] == 'a') {
                update(, n, l, r, );
            } else if(str[] == 'q') {
                int ans = query(, n, l, r, );
                printf("%d\n", ans);
            }
        }
    }
    return ;
}