initial fin isp rep cin 最大值最小值 ble sequence max

Naive Operations

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 502768/502768 K (Java/Others)
Total Submission(s): 1791 Accepted Submission(s): 772

Problem Description 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 $a_l,a_{l+1}...a_r$
2. query l r: query $\sum_{i=l}^r \lfloor a_i / b_i \rfloor$

Input 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.
$1 \leq n,q \leq 100000$, $1 \leq l \leq r \leq n$, there‘re no more than 5 test cases.

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

Sample Input 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

Sample Output 1 1 2 4 4 6

分析：線段樹模板改一改，維護最大值最小值就好了。

#include <iostream>
#include <string>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <algorithm>
#include <vector>
#include <queue>
#include <deque>
#include <map>
#define range(i,a,b) for(auto i=a;i<=b;++i)
#define LL long long
#define itrange(i,a,b) for(auto i=a;i!=b;++i)
#define rerange(i,a,b) for(auto i=a;i>=b;--i)
#define fill(arr,tmp) memset(arr,tmp,sizeof(arr))
using namespace std;
int b[int(1e5+5)],n,q;
template <class T>
class segtree{
private:
    T *add,*cnt,*minb,*maxa;
    void pushup(int rt){
        minb[rt]=min(minb[rt<<1],minb[rt<<1|1]);
        cnt[rt]=cnt[rt<<1]+cnt[rt<<1|1];
        maxa[rt]=max(maxa[rt<<1],maxa[rt<<1|1]);
    }
    void pushdown(int rt){
        if(add[rt]){
            int v=add[rt];
            add[rt]=0;
            maxa[rt<<1]+=v;
            maxa[rt<<1|1]+=v;
            add[rt<<1]+=v;
            add[rt<<1|1]+=v;
        }
    }
public:
    explicit segtree(int len=int(1e5+5)){
        add=new T[len<<2];fill(add,0);
        cnt=new T[len<<2];fill(cnt,0);
        minb=new T[len<<2];fill(minb,0);
        maxa=new T[len<<2];fill(maxa,0);
    }
    void build(int l,int r,int rt){
        add[rt]=0;
        if(l==r){
            cnt[rt]=maxa[rt]=0;
            minb[rt]=b[l];
            return;
        }
        int m=(l+r)>>1;
        build(l,m,rt<<1);
        build(m+1,r,rt<<1|1);
        pushup(rt);
    }
    void update(int L,int R,T c,int l,int r,int rt){
        if(L<=l&&r<=R){
            maxa[rt]++;
            if(maxa[rt]<minb[rt]){
                ++add[rt];
                return;
            }
            if(l==r&&maxa[rt]>=minb[rt]){
                ++cnt[rt];
                minb[rt]+=b[l];
                return;
            }
        }
        pushdown(rt);
        int m=(l+r)>>1;
        if(L<=m)update(L,R,0,l,m,rt<<1);
        if(m<R)update(L,R,0,m+1,r,rt<<1|1);
        pushup(rt);
    }
    T query(int L,int R,int l,int r,int rt){
        if(L<=l&&r<=R)return cnt[rt];
        int m=(l+r)>>1;
        pushdown(rt);
        T ret=0;
        if(L<=m)ret+=query(L,R,l,m,rt<<1);
        if(m<R)ret+=query(L,R,m+1,r,rt<<1|1);
        return ret;
    }
};
segtree<int>tree;
void init(){}
void solve(){
    while(cin>>n>>q){
        range(i,1,n)scanf("%d",b+i);
        tree.build(1,n,1);
        char op[6];int l,r;
        while(q--){
            scanf("%s%d%d",op,&l,&r);
            if(op[0]==‘a‘)tree.update(l,r,0,1,n,1);
            else printf("%d\n",tree.query(l,r,1,n,1));
        }
    }
}
int main() {
    init();
    solve();
    return 0;
}

View Code

HDU 6315: Naive Operations