杭电多校 Naive Operations 线段树+思维

Naive Operations

时间限制: 3 Sec  内存限制: 512 MB
提交: 17  解决: 9
[提交] [状态] [讨论版] [命题人:外部导入]

题目描述

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 

输入

There are multiple test cases, please read till the end of input file.
For each test case, in the first line, two integers n,q, representing the length of a,b and the number of queries.
In the second line, n integers separated by spaces, representing permutation b.
In the following q lines, each line is either in the form 'add l r' or 'query l r', representing an operation.
1≤n,q≤100000, 1≤l≤r≤n, there're no more than 5 test cases.

输出

Output the answer for each 'query', each one line.

样例输入

5 12
1 5 2 4 3
add 1 4
query 1 4
add 2 5
query 2 5
add 3 5
query 1 5
add 2 4
query 1 4
add 2 5
query 2 5
add 2 2
query 1 5

样例输出

1
1
2
4
4
6

[提交][状态]

因为a/b向下取整，所以每次当a加到等于b时，才会产生一个贡献，可以倒过来想，直接减b，减到0就增加1的贡献

用线段树来维护

代码：

#include <bits/stdc++.h>
using namespace std;
const  int maxx=1e5+100;
const int INF=1e9;
const int MOD=1e9+7;
typedef long long ll;
int b[maxx];
struct node
{
    int pos,v;
}mx[maxx<<2];
int lazy[maxx<<2];
void pushup(int rt)
{
    mx[rt].pos=mx[rt<<1].pos+mx[rt<<1|1].pos;
    mx[rt].v=min(mx[rt<<1].v,mx[rt<<1|1].v);
}
void pushdown(int rt)
{
    if(lazy[rt]){
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        mx[rt<<1].v-=lazy[rt];
        mx[rt<<1|1].v-=lazy[rt];
        lazy[rt]=0;
    }
}
void build(int l,int r,int rt)
{
    lazy[rt]=0;
    if(l==r){
        mx[rt].pos=0;
        mx[rt].v=b[l];
        return ;
    }
    int mid=(l+r)>>1;
    build(l,mid,rt<<1);
    build(mid+1,r,rt<<1|1);
    pushup(rt);
}
void change(int l,int r,int rt)
{
    if(l==r){
        if(mx[rt].v==0){
            mx[rt].pos++;
            mx[rt].v=b[l];
        }
        lazy[rt]=0;
        return ;
    }
    pushdown(rt);
    int mid=(l+r)>>1;
    if(mx[rt<<1].v==0)  change(l,mid,rt<<1);
    if(mx[rt<<1|1].v==0)  change(mid+1,r,rt<<1|1);
    pushup(rt);
}
void update(int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R){
        lazy[rt]++;
        mx[rt].v--;
        if(mx[rt].v==0){
            change(l,r,rt);
        }
        return ;
    }
    pushdown(rt);
    int mid=(l+r)>>1;
    if(L<=mid)   update(L,R,l,mid,rt<<1);
    if(R>mid)    update(L,R,mid+1,r,rt<<1|1);
    pushup(rt);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l && r<=R)  return mx[rt].pos;
    pushdown(rt);
    int mid=(l+r)>>1;
    int ret=0;
    if(L<=mid)  ret+=query(L,R,l,mid,rt<<1);
    if(R>mid)   ret+= query(L,R,mid+1,r,rt<<1|1);
    //else        return query(L,R,l,mid,rt<<1)+query(L,R,mid+1,r,rt<<1|1);
    return ret;
}
int main()
{
    int n,q;
    while(~scanf("%d%d",&n,&q)){
        for(int i=1; i<=n; i++){
            scanf("%d",&b[i]);
        }
        build(1,n,1);
        char s[10];
        int l,r;
        while(q--){
            scanf("%s%d%d",s,&l,&r);
            if(s[0]=='a')  update(l,r,1,n,1);
            else           printf("%d\n",query(l,r,1,n,1));
        }
    }
 
    return 0;
}