数列分块入门 4

#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstring>
#include <stdio.h>
#include <algorithm>
#include <map>
#include <queue>
#include <set>
#include <sstream>
#include <vector>
#include <cmath>
#include <stack>
#include <random>
using namespace std;
#define io ios::sync_with_stdio(0),cin.tie(0)
#define ms(arr) memset(arr,0,sizeof(arr))
#define LD long double
#define LL long long
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
#define inf 1<<30
#define  ull unsigned long long
const int Mod = 998244353;
const int maxn = 1e6 + 5;
int read() {
    int x = 0, f = 1;
    char ch = getchar();

    while (ch < '0' || ch > '9') {
        if (ch == '-')
            f = -1;

        ch = getchar();
    }

    while (ch >= '0' && ch <= '9') {
        x = x * 10 + ch - '0';
        ch = getchar();
    }

    return x * f;
}
int n, opt, l, r, c, block;
LL a[maxn];
int bl[maxn];
LL sum[maxn];
LL change[maxn];
void add(int x, int y, int val) {
    for (int i = x; i <= min(block * bl[x], y); i++) {//直接暴力维护x块
        a[i] += val;
        sum[bl[i]] += val;
    }

    if (bl[x] != bl[y]) {//如果不在同一块
        for (int i = block * (bl[y] - 1) + 1; i <= min(block * bl[y], y); i++) {//那y块也暴力维护
            a[i] += val;
            sum[bl[i]] += val;
        }

        for (int i = bl[x] + 1; i < bl[y]; i++) {//x和y块之间直接整块维护
            change[i] += val;
        }
    }
}
void query(int x, int y, int k) {
    LL ans = 0;

    for (int i = x; i <= min(block * bl[x], y); i++) {//加上x块的
        ans = (ans + a[i] + change[bl[i]]) % k;
    }

    if (bl[x] != bl[y]) {//不在同一块
        for (int i = block * (bl[y] - 1) + 1; i <= min(block * bl[y], y); i++) {//加上y块的
            ans = (ans + a[i] + change[bl[i]]) % k;
        }

        for (int i = bl[x] + 1; i < bl[y]; i++) {//加上x和y整块的和
            ans = (ans + change[i] * block % k + sum[i] % k) % k;
        }
    }

    ans = ans % k;
    printf("%lld\n", ans);
}
int main() {
    io;
    n = read();
    block = sqrt(n);

    for (int i = 1; i <= n; i++) {
        a[i] = read();
    }

    for (int i = 1; i <= n; i++) {
        bl[i] = (i - 1) / block + 1;
        sum[bl[i]] += a[i];
    }

    for (int i = 1; i <= n; i++) {
        opt = read();
        l = read();
        r = read();
        c = read();

        if (opt == 0)
            add(l, r, c);
        else
            query(l, r, c + 1);
    }

    return 0;
}