《牛客练习赛28-B》

这题主要就是多了一个平方和的操作。

我们维护平方和的值的时候。

需要注意在下放的时候，要先把乘法之后的sum1算出来，这对算sum1最终的值没有影响。

但是对sum2的值有影响。因为我们在计算中就在更新adtag的值，所以这个adtag它的sum1应该最终化。

 

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
typedef pair<LL,int> pii;
const int N = 1e4+5;
const int M = 1e6+5;
const LL Mod = 998244353;
#define pi acos(-1)
#define INF 1e18
#define CT0 cin.tie(0),cout.tie(0)
#define IO ios::sync_with_stdio(false)
#define dbg(ax) cout << "now this num is " << ax << endl;
namespace FASTIO{
    inline LL read(){
        LL x = 0,f = 1;char c = getchar();
        while(c < '0' || c > '9'){if(c == '-') f = -1;c = getchar();}
        while(c >= '0' && c <= '9'){x = (x<<1)+(x<<3)+(c^48);c = getchar();}
        return x*f;
    }
}
using namespace FASTIO;

LL a[N];
struct Node{LL L,r,adtag,mutag,sum1,sum2;}node[N << 2];
void Pushup(int idx)
{
    node[idx].sum1 = node[idx << 1].sum1 + node[idx << 1 | 1].sum1;
    node[idx].sum2 = node[idx << 1].sum2 + node[idx << 1 | 1].sum2;
}
void Pushdown(int idx)
{
    LL mutag = node[idx].mutag,adtag = node[idx].adtag;
    LL lslen = node[idx << 1].r - node[idx << 1].L + 1;
    LL rslen = node[idx << 1 | 1].r - node[idx << 1 | 1].L + 1;
    
    node[idx << 1].sum1 *= mutag;
    node[idx << 1].sum2 *= mutag * mutag;
    node[idx << 1 | 1].sum1 *= mutag;
    node[idx << 1 | 1].sum2 *= mutag * mutag;

    node[idx << 1].sum2 += 2 * adtag * node[idx << 1].sum1 + adtag * adtag * lslen;
    node[idx << 1 | 1].sum2 += 2 * adtag * node[idx << 1 | 1].sum1 + adtag * adtag * rslen;
    node[idx << 1].sum1 += adtag * lslen;
    node[idx << 1 | 1].sum1 += adtag * rslen;

    node[idx << 1].adtag *= mutag,node[idx << 1].adtag += adtag,node[idx << 1].mutag *= mutag;
    node[idx << 1 | 1].adtag *= mutag,node[idx << 1 | 1].adtag += adtag,node[idx << 1 | 1].mutag *= mutag;
    
    node[idx].mutag = 1,node[idx].adtag = 0;
}
void build(int L,int r,int idx)
{
    node[idx].L = L,node[idx].r = r,node[idx].sum1 = node[idx].sum2 = 0;
    node[idx].adtag = 0,node[idx].mutag = 1;
    if(L == r)
    {
        node[idx].sum1 = a[L];
        node[idx].sum2 = a[L] * a[L];
        return ;
    }
    int mid = (L + r) >> 1;
    build(L,mid,idx << 1);
    build(mid + 1,r,idx << 1 | 1);
    Pushup(idx);
}
void tree_mul(int L,int r,int idx,LL val)
{
    if(node[idx].L >= L && node[idx].r <= r)
    {
        node[idx].sum1 *= val;
        node[idx].sum2 *= val * val;
        node[idx].adtag *= val;
        node[idx].mutag *= val;
        return ;
    }
    int mid = (node[idx].L + node[idx].r) >> 1;
    Pushdown(idx);
    if(mid >= L) tree_mul(L,r,idx << 1,val);
    if(mid < r) tree_mul(L,r,idx << 1 | 1,val);
    Pushup(idx);
}
void tree_add(int L,int r,int idx,LL val)
{
    if(node[idx].L >= L && node[idx].r <= r)
    {
        node[idx].adtag += val;
        node[idx].sum2 += node[idx].sum1 * val * 2 + (node[idx].r - node[idx].L + 1) * val * val;
        node[idx].sum1 += (node[idx].r - node[idx].L + 1) * val;
        return ;
    }
    int mid = (node[idx].L + node[idx].r) >> 1;
    Pushdown(idx);
    if(mid >= L) tree_add(L,r,idx << 1,val);
    if(mid < r) tree_add(L,r,idx << 1 | 1,val);
    Pushup(idx);
}
LL query(int L,int r,int idx,int id)
{
    if(node[idx].L >= L && node[idx].r <= r)
    {
        if(id == 1) return node[idx].sum1;
        else return node[idx].sum2;
    }
    int mid = (node[idx].L + node[idx].r) >> 1;
    LL ans = 0;
    Pushdown(idx);
    if(mid >= L) ans += query(L,r,idx << 1,id);
    if(mid < r) ans += query(L,r,idx << 1 | 1,id);
    return ans;
}
int main()
{
    int n,m;n = read(),m = read();
    for(int i = 1;i <= n;++i) a[i] = read();
    build(1,n,1);
    while(m--)
    {
        int op,x,y;op = read(),x = read(),y = read();
        if(op == 1) printf("%lld\n",query(x,y,1,1));
        else if(op == 2) printf("%lld\n",query(x,y,1,2));
        else if(op == 3)
        {
            LL k;k = read();
            tree_mul(x,y,1,k);
        }
        else
        {
            LL k;k = read();
            tree_add(x,y,1,k);
        }
    }
    return 0;
}