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P5309 [Ynoi2011] 初始化

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P5309 [Ynoi2011] 初始化

题面

题目链接

解法

容易想到对 $x$ 的大小进行分治，如果 $x \le \sqrt{2\times 10^5}$ 就打标记，否则暴力更新。查询也很简单。

AC代码

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/**
 * @file:           P5309.cpp
 * @author:         yaoxi-std
 * @url:            https://www.luogu.com.cn/problem/P5309
*/
// #pragma GCC optimize ("O2")
// #pragma GCC optimize ("Ofast", "inline", "-ffast-math")
// #pragma GCC target ("avx,sse2,sse3,sse4,mmx")
#include <bits/stdc++.h>
using namespace std;
#define resetIO(x) \
    freopen(#x ".in", "r", stdin), freopen(#x ".out", "w", stdout)
#define debug(fmt, ...) \
    fprintf(stderr, "[%s:%d] " fmt "\n", __FILE__, __LINE__, ##__VA_ARGS__)
template <class _Tp>
inline _Tp& read(_Tp& x) {
    bool sign = false; char ch = getchar(); double tmp = 1;
    for (; !isdigit(ch); ch = getchar()) sign |= (ch == '-');
    for (x = 0; isdigit(ch); ch = getchar()) x = x * 10 + (ch ^ 48);
    return sign ? (x = -x) : x;
}
template <class _Tp>
inline void write(_Tp x) {
    if (x < 0) putchar('-'), x = -x;
    if (x > 9) write(x / 10);
    putchar((x % 10) ^ 48);
}
bool m_be;
using ll = long long;
const int MAXN = 2e5 + 10;
const int MAXB = 450;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
int n, q, a[MAXN];
int blk, bn, bl[MAXB], br[MAXB], pos[MAXN];
int bsum[MAXB], psum[MAXN], sum[MAXB][MAXB];
inline void add(int& x, int y) { (x += y), (x >= MOD) && (x -= MOD); }
inline int vadd(int x, int y) { return x += y, x >= MOD ? x - MOD : x; }
bool m_ed;
signed main() {
    // debug("Mem %.5lfMB.", fabs(&m_ed - &m_be) / 1024 / 1024);
    read(n), read(q);
    for (int i = 1; i <= n; ++i) read(a[i]);
    blk = ceil(sqrt(n));
    for (int l = 1; l <= n; l += blk) {
        int r = min(n, l + blk - 1);
        ++bn, bl[bn] = l, br[bn] = r;
        for (int i = l; i <= r; ++i) pos[i] = bn;
    }
    for (int i = 1; i <= n; ++i) add(psum[i], a[i]), add(bsum[pos[i]], a[i]);
    while (q--) {
        int op;
        if (read(op) == 1) {
            int x, y, z; read(x), read(y), read(z), add(z, 0);
            if (x <= blk) {
                for (int i = y; i <= x; ++i) add(sum[x][i], z);
            } else {
                for (int i = y; i <= n; i += x) add(psum[i], z), add(bsum[pos[i]], z);
            }
        } else {
            int l, r, ans = 0; read(l), read(r);
            if (pos[l] == pos[r]) {
                for (int i = l; i <= r; ++i) add(ans, psum[i]);
            } else {
                for (int i = l; i <= br[pos[l]]; ++i) add(ans, psum[i]);
                for (int i = bl[pos[r]]; i <= r; ++i) add(ans, psum[i]);
                for (int i = pos[l] + 1; i < pos[r]; ++i) add(ans, bsum[i]);
            }
            for (int i = 1; i <= blk; ++i) {
                int bl = (l - 1) / i + 1, br = (r - 1) / i + 1;
                int pl = (l - 1) % i + 1, pr = (r - 1) % i + 1;
                add(ans, vadd(sum[i][pr], MOD - sum[i][pl - 1]));
                if (bl != br) add(ans, 1ll * (br - bl) * sum[i][i] % MOD);
            }
            write(ans), putchar('\n');
        }
    }
    return 0;
}