Codechef Chef and Churu

Description

$有一个n个数的数组a，有n个函数，每个函数是返回[l_i,r_i]的和$

有两种操作

1 x y:将数组第x个元素值修改为y

2 m n:询问[m,n]函数的和

$n,q\leq10^5$

Solution

我们可以考虑分块，将函数分为$\sqrt{n}块$，预处理出每块函数和以及每块函数中每个数算的次数，再用一个树状数组求数组的和

修改时根据每块函数中第x个数出现次数更新答案，树状数组单点修改，询问时随便搞搞就行了

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 100010;
int n, q, a[N], L[N], R[N], cnt[1500][N], num[1500][N];
LL c[N], sum[N];
inline int read(int &t) {
    int f = 1;char c;
    while (c = getchar(), c < ‘0‘ || c > ‘9‘) if (c == ‘-‘) f = -1;
    t = c - ‘0‘;
    while (c = getchar(), c >= ‘0‘ && c <= ‘9‘) t = t * 10 + c - ‘0‘;
    t *= f;
}
void add(int i, int x) {
    for (;i <= n; i += i & -i)  c[i] += x;
}
void change(int i, int x) {
    add(i, x - a[i]);
    a[i] = x;
}
LL ask(int i) {
    LL t = 0;
    for (; i; i -= i & -i)  t += c[i];
    return t;
}
int main() {
    read(n);
    int bl = (int)sqrt(n + 0.5);
    int S = n / bl + ((n % bl) ? 1 : 0);
    for (int i = 1; i <= n; ++i)    read(a[i]);
    for (int i = 1; i <= n; ++i)    add(i, a[i]);
    for (int i = 0; i < n; ++i) {
        read(L[i]), read(R[i]);
        ++cnt[i / bl][L[i]];
        --cnt[i / bl][R[i] + 1];
    }
    for (int i = 0; i < S; ++i) 
        for (int j = 1; j <= n; ++j) {
            num[i][j] = cnt[i][j] + num[i][j - 1];
            sum[i] += 1ull * num[i][j] * a[j];
        }
    read(q);
    while (q--) {
        int op, l, r;
        read(op), read(l), read(r);
        if (op == 1) {
            for (int i = 0; i < S; ++i) sum[i] += 1ull * num[i][l] * (r - a[l]);
            change(l, r);
        }
        else {
            --l, --r;
            LL ans = 0;
            int x = l / bl, y = r / bl;
            if (x == y) for (int i = l; i <= r; ++i)    ans += ask(R[i]) - ask(L[i] - 1);
            else{
                x = (l % bl ? x + 1 : x), y = (r + 1) % bl ? y - 1 : y;
                for (int i = x; i <= y; ++i)    ans += sum[i];
                while (l % bl)  ans += ask(R[l]) - ask(L[l++] - 1);
                while ((r + 1) % bl)    ans += ask(R[r]) - ask(L[r--] - 1);
            }
            printf("%llu\n", ans);
        }
    }
    return 0;
}