BZOJ4373: 算术天才⑨与等差数列(线段树 hash?)

题意

题目链接

sol

正经做法不会，听lxl讲了一种很神奇的方法

我们考虑如果满足条件，那么需要具备什么条件

设mx为询问区间最大值，mn为询问区间最小值

1. mx - mn = (r - l) * k

2. 区间和 = mn * len + \(\frac{n * (n - 1)}{2} k\)

3. \(\text{立方和} = \sum_{i = 1}^{len} (mn + (i - 1)k) ^2\)

第三条后面的可以直接推式子推出来(\(\sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}\))

最后的/6可以直接乘一下然后ull自然溢出。

#include<bits/stdc++.h> 
#define ull unsigned long long 
#define ll long long 
#define fin(x) {freopen(#x".in","r",stdin);}
#define fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int maxn = 4e6 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m;
#define ls k << 1
#define rs k << 1 | 1
struct {
    int l, r, mn, mx;
    ull s, s2;
}t[maxn];
void update(int k) {
    t[k].mn = min(t[ls].mn, t[rs].mn);
    t[k].mx = max(t[ls].mx, t[rs].mx);
    t[k].s  = t[ls].s + t[rs].s;
    t[k].s2 = t[ls].s2 + t[rs].s2;
}
void build(int k, int ll, int rr) {
    t[k].l = ll; t[k].r = rr;
    if(ll == rr) {t[k].mn = t[k].mx = t[k].s = read(); t[k].s2 = t[k].s * t[k].s ; return ;}
    int mid = ll + rr >> 1;
    build(ls, ll, mid); build(rs, mid + 1, rr);
    update(k);
}
void modify(int k, int p, int v) {
    if(t[k].l == t[k].r) {t[k].mn = t[k].mx = t[k].s = v; t[k].s2 = t[k].s * t[k].s; return ;}
    int mid = t[k].l + t[k].r >> 1;
    if(p <= mid) modify(ls, p, v);
    if(p  > mid) modify(rs, p, v);
    update(k);
}
int querymn(int k, int ql, int qr) {
    if(ql <= t[k].l && t[k].r <= qr) return t[k].mn;
    int mid = t[k].l + t[k].r >> 1;
    if(ql > mid) return querymn(rs, ql, qr);
    else if(qr <= mid) return querymn(ls, ql, qr);
    else return min(querymn(ls, ql, qr), querymn(rs, ql, qr));
}
int querymx(int k, int ql, int qr) {
    if(ql <= t[k].l && t[k].r <= qr) return t[k].mx;
    int mid = t[k].l + t[k].r >> 1;
    if(ql > mid) return querymx(rs, ql, qr);
    else if(qr <= mid) return querymx(ls, ql, qr);
    else return max(querymx(ls, ql, qr), querymx(rs, ql, qr));
}
ull querysum(int k, int ql, int qr) {
    if(ql <= t[k].l && t[k].r <= qr) return t[k].s;
    int mid = t[k].l + t[k].r >> 1;
    if(ql > mid) return querysum(rs, ql, qr);
    else if(qr <= mid) return querysum(ls, ql, qr);
    else return querysum(ls, ql, qr) + querysum(rs, ql, qr);
}
ull querysum2(int k, int ql, int qr) {
    if(ql <= t[k].l && t[k].r <= qr) return t[k].s2;
    int mid = t[k].l + t[k].r >> 1;
    if(ql > mid) return querysum2(rs, ql, qr);
    else if(qr <= mid) return querysum2(ls, ql, qr);
    else return querysum2(ls, ql, qr) + querysum2(rs, ql, qr);
}
signed main() {
    n = read(); m = read();
    build(1, 1, n); 
    int gg = 0;
    while(m--) {
        int opt = read();
        if(opt == 1) {
            int x = read() ^ gg, y = read() ^ gg;
            modify(1, x, y);
        } else {
            int l = read() ^ gg, r = read() ^ gg; ull k = read() ^ gg;
            ull n = r - l + 1, mn = querymn(1, l, r), mx = querymx(1, l, r);
            ull s = querysum(1, l, r), s2 = querysum2(1, l, r), ns = mn * n + n * (n - 1) * k / 2;
            ull gg = (6 * mn * mn * n) + (6 * mn * k * n * (n - 1))  + (k * k * (n - 1) * n * (2 * (n - 1) + 1)), 
                gg2 = 6 * s2;
            if((mx - mn == (r - l) * k) && 
               (mn * n + n * (n - 1) / 2 * k == s) &&
               (gg  == gg2)) puts("yes"), gg++;
            else puts("no");
        }
    }
    return 0;
}
/*
5 3
1 3 2 5 6
2 2 2 23333
1 5 4
2 1 5 1
*/