c/c++语言开发共享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 */