线段树 B数据结构 牛客练习赛28

链接：https://ac.nowcoder.com/acm/contest/200/B
来源：牛客网

题目描述

qn姐姐最好了~

    qn姐姐给你了一个长度为n的序列还有m次操作让你玩，

    1 l r 询问区间[l,r]内的元素和

    2 l r 询问区间[l,r]内的元素的平方 和

    3 l r x 将区间[l,r]内的每一个元素都乘上x

    4 l r x 将区间[l,r]内的每一个元素都加上x

输入描述:

第一行两个数n,m

接下来一行n个数表示初始序列

就下来m行每行第一个数为操作方法opt，

若opt=1或者opt=2，则之后跟着两个数为l,r

若opt=3或者opt=4，则之后跟着三个数为l,r,x

操作意思为题目描述里说的

输出描述:

对于每一个操作1,2，输出一行表示答案

示例1

输入

复制

5 6
1 2 3 4 5
1 1 5
2 1 5
3 1 2 1
4 1 3 2
1 1 4
2 2 3

输出

复制

15
55
16
41

备注:

对于100%的数据 n=10000,m=200000 (注意是等于号)

保证所有询问的答案在long long 范围内

这个比较简单，但是出现了一个很难找的bug。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <queue>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#define inf 0x3f3f3f3f
using namespace std;
typedef long long ll;
const int maxn = 2e5 + ;

ll a[maxn];

struct node
{
    int l, r;
    ll lazyc, lazyadd;
    ll sum, all;
}tree[maxn*];

void push_up(int id)
{
    tree[id].sum = tree[id << ].sum + tree[id <<  | ].sum;
    tree[id].all = tree[id << ].all + tree[id <<  | ].all;
}

void push_down(int id)
{
    if(tree[id].lazyc>)
    {
        tree[id << ].sum = tree[id << ].sum*tree[id].lazyc;
        tree[id <<  | ].sum = tree[id <<  | ].sum*tree[id].lazyc;
        tree[id << ].all = tree[id << ].all*tree[id].lazyc*tree[id].lazyc;
        tree[id <<  | ].all = tree[id <<  | ].all*tree[id].lazyc*tree[id].lazyc;
        tree[id << ].lazyc *= tree[id].lazyc;
        tree[id <<  | ].lazyc *= tree[id].lazyc;
        tree[id].lazyc = ;
    }
    if(tree[id].lazyadd)
    {
        tree[id << ].all += tree[id << ].sum * *tree[id].lazyadd + (tree[id << ].r - tree[id << ].l + )*tree[id].lazyadd*tree[id].lazyadd;
        tree[id <<  | ].all += tree[id <<  | ].sum *  *tree[id].lazyadd+ (tree[id <<  | ].r - tree[id <<  | ].l + )*tree[id].lazyadd*tree[id].lazyadd;
        tree[id << ].sum += (tree[id<<].r - tree[id<<].l + )*tree[id].lazyadd;
        tree[id <<  | ].sum += (tree[id <<  | ].r - tree[id <<  | ].l + )*tree[id].lazyadd;
        tree[id << ].lazyadd += tree[id].lazyadd;
        tree[id <<  | ].lazyadd += tree[id].lazyadd;
        tree[id].lazyadd = ;
    }
}

void build(int id,int l,int r)
{
    tree[id].l = l;
    tree[id].r = r;
    tree[id].lazyc = ;
    tree[id].lazyadd = ;
    if(l==r)
    {
        tree[id].sum = a[l];
        tree[id].all = a[l] * a[l];
        return;
    }
    int mid = (l + r) >> ;
    build(id << , l, mid);
    build(id <<  | , mid + , r);
    push_up(id);
}

void updatec(int id,int l,int r,int x)
{
    push_down(id);
    if(l<=tree[id].l&&r>=tree[id].r)
    {
        tree[id].lazyc *= x;
        tree[id].lazyadd *= x;
        tree[id].sum = tree[id].sum*x;
        tree[id].all = tree[id].all*x*x;
        return;
    }
    int mid = (tree[id].l + tree[id].r) >> ;
    if (l <= mid) updatec(id << , l, r, x);
    if (r > mid) updatec(id <<  | , l, r, x);
    push_up(id);
}

void updateadd(int id,int l,int r,int x)
{
    push_down(id);
    if(l<=tree[id].l&&r>=tree[id].r)
    {
        tree[id].lazyadd += x;
        tree[id].all += tree[id].sum * *x + (tree[id].r - tree[id].l + )*x*x;
        tree[id].sum += (tree[id].r - tree[id].l + )*x;
        return;
    }
    int mid = (tree[id].l + tree[id].r) >> ;
    if (l <= mid) updateadd(id << , l, r, x);
    if (r > mid) updateadd(id <<  | , l, r, x);//这里的id<<1|1忘记+1了，就写成了id<<1
    push_up(id);
}

ll querysum(int id,int l,int r)
{

    if(l<=tree[id].l&&r>=tree[id].r)
    {
        return tree[id].sum;
    }
    ll ans = ;
    push_down(id);
    int mid = (tree[id].l + tree[id].r) >> ;
    if (l <= mid) ans += querysum(id << , l, r);
    if (r > mid) ans += querysum(id <<  | , l, r);
    return ans;
}

ll queryc(int id,int l,int r)
{
    if(l<=tree[id].l&&r>=tree[id].r)
    {
        return tree[id].all;
    }
    push_down(id);
    ll ans = ;
    int mid = (tree[id].l + tree[id].r) >> ;
    if (l <= mid) ans += queryc(id << , l, r);
    if (r > mid) ans += queryc(id <<  | , l, r);
    return ans;
}

int main()
{
    int n, m;
    scanf("%d%d", &n, &m);
    for (int i = ; i <= n; i++) scanf("%lld", &a[i]);
    build(, , n);
    int opt, l, r, x;
    for(int i=;i<=m;i++)
    {
        scanf("%d%d%d", &opt, &l, &r);
        if(opt==)
        {
            ll ans = querysum(, l, r);
            printf("%lld\n", ans);
        }
        if(opt==)
        {
            ll ans = queryc(, l, r);
            printf("%lld\n", ans);
        }
        if(opt==)
        {
            scanf("%d", &x);
            updatec(, l, r, x);
        }
        if(opt==)
        {
            scanf("%d", &x);
            updateadd(, l, r, x);
        }
    }
    return ;
}