題意簡述

已知一個數列，你需要進行下面兩種操作：
1.將某區間每一個數加上x
2.求出某區間每一個數的和

題解思路

對于一個長度為n的序列，我們可以講其中的元素分為\( \sqrt{n} \) 個連續的子序列，每塊的長度自然就為\( \sqrt{n} \)。
我們更新一段區間[l,r]，可以先更新l到l所在塊的右端點，r到r所在塊的右端點到r和中間的整個區間。

代碼

#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\n", query(x, y));}}return 0; }

轉載于:https://www.cnblogs.com/xuyixuan/p/9429524.html

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