模板分块

Posted 2020-12-28 xuyixuan

题意简述

已知一个数列，你需要进行下面两种操作：
1.将某区间每一个数加上x
2.求出某区间每一个数的和

题解思路

对于一个长度为n的序列，我们可以讲其中的元素分为$ sqrt{n}(个连续的子序列，每块的长度自然就为) sqrt{n}$。
我们更新一段区间[l,r]，可以先更新l到l所在块的右端点，r到r所在块的右端点到r和中间的整个区间。

代码

#include <cmath>
#include <cstdio>
using namespace std;
typedef long long ll;
struct Point{
    ll w, num;
};
struct K{
    ll l, r, tot, sum;
    ll len()
    {
        return r - l + 1;
    } 
};
ll n, m, s, len;
Point p[100001];
K k[501];
void add(ll x, ll y, ll t)
{
    if (p[x].num == p[y].num)
    {
        for (register ll i = x; i <= y; ++i)
            p[i].w += t;
        k[p[x].num].tot += (y - x + 1) * t;
        return;
    }
    for (register ll i = x; i <= k[p[x].num].r; ++i)
        p[i].w += t;
    k[p[x].num].tot += (k[p[x].num].r - x + 1) * t;
    for (register ll i = k[p[y].num].l; i <= y; ++i)
        p[i].w += t;
    k[p[y].num].tot += (y - k[p[y].num].l + 1) * t;
    for (register ll i = p[x].num + 1; i <= p[y].num - 1; ++i)
    {
        k[i].tot += t * k[i].len();
        k[i].sum += t;
    }
}
ll query(ll x, ll y)
{
    ll ans = 0;
    if (p[x].num == p[y].num)
    {
        for (register ll i = x; i <= y; ++i)
            ans += p[i].w;
        return ans;
    }
    for (register ll i = x; i <= k[p[x].num].r; ++i)
        ans += p[i].w + k[p[x].num].sum;
    for (register ll i = k[p[y].num].l; i <= y; ++i)
        ans += p[i].w + k[p[y].num].sum;
    for (register ll i = p[x].num + 1; i <= p[y].num - 1; ++i)
        ans += k[i].tot;
    return ans;
}
int main()
{
    scanf("%lld%lld", &n, &m);
    len = sqrt(n);
    s = n / len + (bool)(n % len);
    for (register ll i = 1; i <= s; ++i)
    {
        k[i].l = (i - 1) * len + 1;
        k[i].r = i * len;
    }
    k[s].r = n;
    for (register ll i = 1; i <= n; ++i)
    {
        scanf("%lld", &p[i].w);
        p[i].num = (i - 1) / len + 1;
        k[p[i].num].tot += p[i].w;
    }
    for (register ll i = 1; i <= m; ++i)
    {
        ll op, x, y, t;
        scanf("%lld", &op);
        if (op == 1)
        {
            scanf("%lld%lld%lld", &x, &y, &t);
            add(x, y, t);
        }
        else
        {
            scanf("%lld%lld", &x, &y);
            printf("%lld
", query(x, y));
        }
    }
    return 0;
}