Luogu P4868 Preprefix sum

题目链接：​​传送门​​

线段树维护前缀和
简单明了
修改就修改
当然还有更快的树状数组差分的做法

 */
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <complex>
#include <algorithm>
#include <climits>
#include <queue>
#include <map>
#include <set>
#include <vector>
#include <iomanip>
#define
#define

using namespace std;
typedef long long ll;
struct node {
    int l, r; ll w, f;
}tree[A];
ll s[A], delta;
int n, m, a[A], x, y;
void build(int k, int l, int r) {
    tree[k].l = l; tree[k].r = r;
    if (l == r) {
        tree[k].w = s[l];
        return;
    }
    int m = (l + r) >> 1;
    build(k << 1, l, m);
    build(k << 1 | 1, m + 1, r);
    tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w;
}
void down(int k) {
    tree[k << 1].f += tree[k].f; tree[k << 1 | 1].f += tree[k].f;
    tree[k << 1].w += tree[k].f * (tree[k << 1].r - tree[k << 1].l + 1);
    tree[k << 1 | 1].w += tree[k].f * (tree[k << 1 | 1].r - tree[k << 1 | 1].l + 1);
    tree[k].f = 0;
}
void change(int k, int l, int r) {
    if (tree[k].l >= l and tree[k].r <= r) {
        tree[k].w += delta * ll(tree[k].r - tree[k].l + 1);
        tree[k].f += delta;
        return;
    }
    if (tree[k].f) down(k);
    int m = (tree[k].l + tree[k].r) >> 1;
    if (l <= m) change(k << 1, l, r);
    if (r > m) change(k << 1 | 1, l, r);
    tree[k].w = tree[k << 1].w + tree[k << 1 | 1].w;
}
ll ask(int k, int l, int r) {
    if (tree[k].l >= l and tree[k].r <= r) return tree[k].w;
    if (tree[k].f) down(k);
    int m = (tree[k].l + tree[k].r) >> 1; ll ans = 0;
    if (l <= m) ans += ask(k << 1, l, r);
    if (r > m) ans += ask(k << 1 | 1, l, r);
    return ans;
}

int main(int argc, char const *argv[]) {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]), s[i] = s[i - 1] + a[i];
    build(1, 1, n);
    while (m--) {
        char opt[7]; scanf("%s", opt);
        if (opt[0] == 'Q') scanf("%d", &x), printf("%lld\n", ask(1, 1, x));
        else scanf("%d%d", &x, &y), delta = y - a[x], a[x] = y, change(1, x, n);
    }
    return 0;
}