HDU 5828 Rikka with Sequence

题意： 对一个长度为10W 的数组进行如下三种操作：

1，区间[l,r]的数都加上x;

2，区间内的没个数都开根号（向下取整）；

3，求区间[l,r]每个元素的和。

官方题解：



但是官方标程也被卡T了。据说hack数据是：10万个2，3，2，3，2，3.......,10万个操作 加6，开根。但是题的确不错。

在本题中我并没有按照官方题解的方法。我们可以想下，发现区间的最大值，最小值的差值会随着开方的次数越来越小。再加上又是区间内的数都加上一个值。

那么我们可以只需要考虑区间内极差小于2的区间即可。刚好我们需要维护区间最大值，最小值。

那么当区间的极差为0时，表示区间内的数都相同，直接开方，该区间的值都覆盖为sqrt(maxx); 区间的最大值，最小值，和，最大值的数量，最小值的数量自然就出来了

当区间的极差为1时，很显然最大值，最小值，及其它们的数量就表示出了区间的所有数。最大值最小值，自然也出来了。

a[i]<10W 那么对整个区间内的每个数开方的次数就很少了。 当开了一定次方之后整个区间内的数的差值就很接近了。

代码如下：

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<vector>
#include<string.h>
#include<map>
#include<queue>
#include<stack>
#include<string>
#include<stdlib.h>
#include<time.h>
#include<set>
#include<math.h>
using namespace std;
typedef long long ll;
const int nn = 600100;
const int inf = (1 << 31);
const ll mod = 998244353;
const ll g = 3;//mod的原根，当且仅当 g^(MOD-1) = 1 % MOD
#define pb push_back
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1

struct FastIO {
static const int S = 1310720;
int wpos; char wbuf[S];
FastIO() : wpos(0) {}
inline int xchar() {
static char buf[S];
static int len = 0, pos = 0;
if (pos == len)
pos = 0, len = fread(buf, 1, S, stdin);
if (pos == len) return -1;
return buf[pos++];
}
inline int xuint() {
int c = xchar(), x = 0;
while (c <= 32) c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x;
}
inline int xint() {
int s = 1, c = xchar(), x = 0;
while (c <= 32) c = xchar();
if (c == '-') s = -1, c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x * s;
}
inline ll xll() {
ll s = 1, c = xchar(), x = 0;
while (c <= 32) c = xchar();
if (c == '-') s = -1, c = xchar();
for (; '0' <= c && c <= '9'; c = xchar()) x = x * 10 + c - '0';
return x * s;
}
inline void xstring(char *s) {
int c = xchar();
while (c <= 32) c = xchar();
for (; c > 32; c = xchar()) *s++ = c;
*s = 0;
}
inline void wchar(int x) {
if (wpos == S) fwrite(wbuf, 1, S, stdout), wpos = 0;
wbuf[wpos++] = x;
}
inline void wint(int x) {
if (x < 0) wchar('-'), x = -x;

char s[24];
int n = 0;
while (x || !n) s[n++] = '0' + x % 10, x /= 10;
while (n--) wchar(s
);
}
inline void wstring(const char *s) {
while (*s) wchar(*s++);
}
~FastIO() {
if (wpos) fwrite(wbuf, 1, wpos, stdout), wpos = 0;
}
} io;
struct node
{
ll minn, maxx,sum;//区间最小值，最大值，区间和
ll maxnum, minnum;//区间最大值，最小值的数量
ll all;//区间覆盖为all,0表示没覆盖
ll lazy;//区间内的数都加上lazy；
};
int n, m;
node t[nn << 2];

void pushup(int rt)
{
t[rt].sum = t[rt << 1].sum + t[rt << 1 | 1].sum;
t[rt].maxx = max(t[rt << 1].maxx, t[rt << 1 | 1].maxx);
t[rt].minn = min(t[rt << 1].minn, t[rt << 1 | 1].minn);
t[rt].maxnum = t[rt].minnum = 0;
if (t[rt].maxx == t[rt << 1].maxx) t[rt].maxnum += t[rt << 1].maxnum;
if (t[rt].maxx == t[rt << 1 | 1].maxx) t[rt].maxnum += t[rt << 1 | 1].maxnum;
if (t[rt].minn == t[rt << 1].minn) t[rt].minnum += t[rt << 1].minnum;
if (t[rt].minn == t[rt << 1 | 1].minn) t[rt].minnum += t[rt << 1 | 1].minnum;
}

void pushdown(int rt,int l,int r)
{
int mid = (l + r) >> 1;
if (t[rt].all)//区间覆盖
{
t[rt << 1].all = t[rt << 1 | 1].all = t[rt].all;
t[rt << 1].sum = (ll)(mid - l + 1)*t[rt].all;
t[rt << 1 | 1].sum = (ll)(r - mid)*t[rt].all;
t[rt << 1].minn = t[rt << 1].maxx = t[rt << 1 | 1].maxx = t[rt << 1 | 1].minn = t[rt].all;
t[rt << 1].minnum = t[rt << 1].maxnum = (ll)(mid - l + 1);
t[rt << 1 | 1].maxnum = t[rt << 1 | 1].minnum = (ll)(r - mid);
t[rt << 1].lazy = t[rt << 1 | 1].lazy = 0;
t[rt].all = 0;
}
if (t[rt].lazy)//区间+lazy
{
t[rt << 1].lazy += t[rt].lazy;
t[rt << 1 | 1].lazy += t[rt].lazy;
t[rt << 1].sum += (ll)(mid - l + 1)*t[rt].lazy;
t[rt << 1 | 1].sum += (ll)(r - mid)*t[rt].lazy;
t[rt << 1].minn += t[rt].lazy;
t[rt << 1 | 1].minn += t[rt].lazy;
t[rt << 1].maxx += t[rt].lazy;
t[rt << 1 | 1].maxx += t[rt].lazy;
t[rt].lazy = 0;
}
}

void build(int l, int r, int rt)
{
t[rt].all = t[rt].lazy = 0;
if (l == r)
{
t[rt].sum = t[rt].maxx = t[rt].minn = io.xll();
t[rt].maxnum = t[rt].minnum = 1;
return;
}
int mid = (l + r) >> 1;
build(lson);
build(rson);
pushup(rt);
}

void add(int l, int r, int rt, int L, int R, ll x)
{
if (L <= l && r <= R)
{
t[rt].sum += (ll)(r - l + 1)*x;
t[rt].maxx += x;
t[rt].minn += x;
t[rt].lazy += x;
return;
}
pushdown(rt, l, r);
int mid = (l + r) >> 1;
if (L <= mid) add(lson, L, R, x);
if (R > mid) add(rson, L, R, x);
pushup(rt);
}
void t_sqrt(int l, int r, int rt, int L, int R)
{
if (L <= l && r <= R && (t[rt].maxx - t[rt].minn) <= 1)
{
if (t[rt].maxx == 1) return;
ll x = t[rt].maxx;
if (t[rt].maxx == t[rt].minn)
{
t[rt].minn = t[rt].maxx = floor(sqrt(x));
t[rt].lazy += t[rt].maxx-x;
t[rt].sum = (ll)(r - l + 1)*t[rt].maxx;
}
else
{
t[rt].maxx = floor(sqrt(t[rt].maxx));
t[rt].minn = floor(sqrt(t[rt].minn));
if (t[rt].maxx == t[rt].minn)
{
t[rt].maxnum = t[rt].minnum = (ll)(r - l + 1);
t[rt].lazy = 0;
t[rt].all = t[rt].maxx;
t[rt].sum = t[rt].maxx*t[rt].maxnum;
}
else
{
t[rt].lazy += t[rt].maxx - x;
t[rt].sum = t[rt].maxx*t[rt].maxnum + t[rt].minn*t[rt].minnum;
}
}
return;
}
int mid = (l + r) >> 1;
pushdown(rt, l, r);
if (L <= mid) t_sqrt(lson, L, R);
if (R > mid) t_sqrt(rson, L, R);
pushup(rt);
}

ll query(int l, int r, int rt, int L, int R)
{
if (L <= l && r <= R)
{
return t[rt].sum;
}
int mid = (l + r) >> 1;
pushdown(rt, l, r);
ll ans = 0;
if (L <= mid) ans += query(lson, L, R);
if (R > mid) ans += query(rson, L, R);
return ans;
}
int main()
{
int t = io.xint();
while (t--)
{
n = io.xll();
m = io.xll();
build(1,n,1);
while (m--)
{
int op = io.xint();
int l = io.xint();
int r = io.xint();
if (op == 1)
{
ll x = io.xll();
add(1, n, 1, l, r, x);
}
else if (op == 2)
{
t_sqrt(1, n, 1, l, r);
}
else
{
printf("%lld\n", query(1, n, 1, l, r));
}
}
}
return 0;
}

加不加输入外挂都是能过的。但是必须要交G++ 不然就算加了输入外挂一样TLE