算法1

线段树标记永久化

时间复杂度

参考文献

C++ 代码

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

typedef long long LL;

const int N = 100010;

struct Node{
    int l, r;
    LL sum, add;
}tr[N * 4];

LL a[N];

int n, m;

LL intersaction(int al, int ar, int bl, int br)
{
    return min(ar, br) - max(al, bl) + 1;
}

void pushup(int u)
{
    tr[u].sum = tr[u << 1].sum + tr[u << 1 | 1].sum; 
}

void build(int u, int l, int r)
{
    tr[u].l = l, tr[u].r = r, tr[u].sum = 0, tr[u].add = 0;
    if(l == r) 
    {
        tr[u].sum = a[l];
        return;
    }
    int mid = (l + r) / 2;
    build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
    pushup(u);
}

LL query(int u, int l, int r, LL add)
{
     if(l <= tr[u].l && r >= tr[u].r) return tr[u].sum + add * intersaction(l, r, tr[u].l, tr[u].r);
     add += tr[u].add;
     int mid = (tr[u].l + tr[u].r) / 2;
     LL res = 0;
     if(l <= mid) res += query(u << 1, l, r, add);
     if(r > mid) res += query(u << 1 | 1, l, r, add);
     return res;
}

void modify(int u, int l, int r, LL d)
{
    tr[u].sum += intersaction(l, r, tr[u].l, tr[u].r) * d;
    if(tr[u].l >= l && tr[u].r <= r)
    {
        tr[u].add += d;
        return;
    }
    int mid = (tr[u].l + tr[u].r) / 2;
    if(l <= mid) modify(u << 1, l, r, d);
    if(r > mid) modify(u << 1 | 1, l, r, d);
}

int main()
{
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++ i) scanf("%lld", &a[i]);
    build(1, 1, n);
    while(m --)
    {
        char op[2];
        scanf("%s", op);
        if(*op == 'Q')
        {
            int l, r;
            scanf("%d%d", &l, &r);
            printf("%lld\n", query(1, l, r, 0ll));
        }
        else
        {
            int l, r;
            LL d;
            scanf("%d%d%lld", &l, &r, &d);
            modify(1, l, r, d);
        }
    }
    return 0;
}