Codechef Chef and church u

Codechef Chef and church u
Description There is an array of n numbers a and there are n functions. Each function returns the sum of [li, ri] There are two types of operations: 1 x y: Change the x element value of the array to y2 m n: Ask the sum of the [m, n] Function N, q ≤ 105 Solution we can consider dividing functions into blocks N ?? √ Block , Pre-process the number of times each function and each number in each function, and then use a tree array to obtain and modify the number of times the number of x appears in each function to update the answer, tree array single-point modification. You just need to Code it when asking.

#include 
  
   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;}