题目描述

给出序列 a1，a2，…an（0≤ai≤109），有关序列的四种操作：

1. al，al+1，…，ar（1≤l≤r≤n）加上 x（-103≤x≤103）

2. al，al+1，…，ar（1≤l≤r≤n）乘上 x（-103≤x≤103）

3. al，al+1，…，ar（1≤l≤r≤n）变成 x－al，x－al+1 ，…，x－ar（-103≤x≤103）

4. 求：

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" alt="" />

（即求：al＋al+1＋al+2＋…＋ar（1≤l≤r≤n）

所有操作均模 109＋7

输入格式

第一行包含两个数 n（1≤n≤105）和 m（1≤m≤105），表示序列长度和操作次数

接下来一行包含 n 个数，空格隔开，表示 a1，a2，…，an 。

接下来 m 行，每行为以下 4 种格式之一：

- 0 l r x ，表示 al，al+1，…，ar 加上 x
- 1 l r x ，表示 al，al+1，…，ar 乘上 x
- 2 l r x ，表示 al，al+1，…，ar（1≤l≤r≤n）变成x－al，x－al+1 ，…，x－ar 。
- 3 l r ，求：

输出格式

对于每次询问，输出单独一行表示答案。

样例数据 1

输入

5 4
1 2 3 4 5
0 1 5 1
1 1 5 -1
2 1 5 1
3 5 5

输出

7

题目分析

同样是线段树裸题。维护区间加、乘标记，对于操作3，先乘-1，再加上x。

坑点同样是下标下放顺序：先乘后加，乘的时候add标记可以直接乘，而加的时候乘号标记必须已经下放。

code

#include<iostream>

#include<cstring>

#include<string>

#include<cstdio>

#include<cstdlib>

#include<algorithm>

#include<cmath>

#include<vector>

using namespace std;

const int N = 1e5 + , Mod = 1e9 + ;

typedef long long ll;

int n, m, data[N];

struct node{

    int len;

    ll sum, add, time;

    node():sum(), time(), len(), add(){}

};

namespace SegTree{

    node tr[N << ];

    inline void upt(int k){

        tr[k].sum = (tr[k << ].sum + tr[k <<  | ].sum) % Mod;

    }

    inline void build(int k, int l, int r){

        tr[k].len = r - l + ;

        if(l == r){

            tr[k].sum = data[l];

            tr[k].add = , tr[k].time = ;

            return;

        }

        int mid = l + r >> , lc = k << , rc = k <<  | ;

        build(lc, l, mid);

        build(rc, mid + , r);

        upt(k);

    }

    inline void Add(int , int);

    inline void plus(ll &x, ll v){

        x = (x + v) % Mod;

        while(x < ) x += Mod;

    }

    inline void mul(ll &x, ll v){

        x = (x * v) % Mod;

        while(x < ) x += Mod;

    }

    inline void Time(int k, int v){

        mul(tr[k].add , v);

        mul(tr[k].sum , v);

        mul(tr[k].time , v);

    }

    inline void Add(int k, int v){

        plus(tr[k].sum, 1LL * tr[k].len * v);

        plus(tr[k].add, v);

    }

    inline void pushDown(int k){

        int lc = k << , rc = k <<  | ;

        if(tr[k].time != ){

            if(tr[k].len > )

                Time(lc, tr[k].time),

                Time(rc, tr[k].time);

            tr[k].time = ;

        }

        if(tr[k].add){

            if(tr[k].len > )

                Add(lc, tr[k].add),

                Add(rc, tr[k].add);

            tr[k].add = ;

        }

    }

    inline void modify(int k, int l, int r, int x, int y, int type, int v){

        pushDown(k);

        if(x <= l && r <= y){

            switch(type){

                case : Add(k, v); break;

                case : Time(k, v); break;

                case : Time(k, -), Add(k, v); break;

            }

            return;

        }

        int mid = l + r >> , lc = k << , rc = k <<  | ;

        if(x <= mid) modify(lc, l, mid, x, y, type, v);

        if(y > mid) modify(rc, mid + , r, x, y, type, v);

        upt(k);

    }

    inline int query(int k, int l, int r, int x, int y){

        pushDown(k);

        if(x <= l && r <= y) return tr[k].sum;

        int mid = l + r >> , lc = k << , rc = k <<  | , ret = ;

        if(x <= mid) ret = (ret + query(lc, l, mid, x, y)) % Mod;

        if(y > mid) ret = (ret + query(rc, mid + , r, x, y)) % Mod;

        return ret;

    }

}using namespace SegTree;

inline int read(){

    int i = , f = ; char ch = getchar();

    for(; (ch < '' || ch > '') && ch != '-'; ch = getchar());

    if(ch == '-') f = -, ch = getchar();

    for(; ch >= '' && ch <= ''; ch = getchar())

        i = (i << ) + (i << ) + (ch - '');

    return i * f;

}

inline void wr(int x){

    if(x < ) putchar('-'), x = -x;

    if(x > ) wr(x / );

    putchar(x %  + '');

}

int main(){

    n = read(), m = read();

    for(int i = ; i <= n; i++) data[i] = read();

    build(, , n);

    for(int i = ; i <= m; i++){

        int opt = read() + , a = read(), b = read(), c;

        if(opt ==  || opt ==  || opt == )

            c = read(), modify(, , n, a, b, opt, c);

        else wr(query(, , n, a, b)), putchar('\n');

    }

    return ;

}

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【序列操作III】线段树