P6638 「JYLOI Round 1」常规 题解

合集 - 题解(97)

1.BZOJ3364 Distance Queries 距离咨询 题解2023-08-082.UVA12390 Distributing Ballot Boxes 题解2023-08-243.UVA1108 Mining Your Own Business 题解2023-08-244.CF1425F Flamingoes of Mystery 题解2023-09-295.SP16113 SUBTLEBA - Trucks Transportation 题解2023-10-016.UVA12655 Trucks 题解2023-10-017.P5943 [POI2002] 最大的园地 题解2023-10-018.SP9494 ZSUM - Just Add It 题解2023-10-029.CF1878C Vasilije in Cacak 题解2023-10-0210.CF1010C Border 题解2023-10-0611.CF131D Subway 题解2023-10-0612.B3610 [图论与代数结构 801] 无向图的块 题解2023-11-0413.P9801 [NERC2018] King Kog’s Reception2023-11-0514.CF1089K King Kog's Reception 题解2023-11-0515.SP28304 ADATEAMS - Ada and Teams 题解2023-11-1916.CF327C Magic Five 题解2023-12-1617.P1405 苦恼的小明 题解2023-12-1618.SP10050 POWTOW - Power Tower City 题解2023-12-1619.SP21690 POWERUP - Power the Power Up 题解2023-12-1620.P2898 [USACO08JAN] Haybale Guessing G 题解2023-12-3121.AT_arc125_c [ARC125C] LIS to Original Sequence 题解2024-01-1322.CF1433E Two Round Dances 题解2024-01-2723.CF739A Alyona and mex 题解2024-01-2724.AT_abc270_g [ABC270G] Sequence in mod P 题解2024-02-0725.SP277 CTGAME - City Game 题解2024-02-0726.P8670 [蓝桥杯 2018 国 B] 矩阵求和 题解2024-02-1027.UVA12467 Secret Word 题解2024-02-1328.SP34020 ADAPET - Ada and Pet 题解2024-02-1329.AT_joig2021_d 展覧会 2 (Exhibition 2) 题解2024-03-0330.AT_abc184_f [ABC184F] Programming Contest 题解2024-03-0331.AT_abc343_g [ABC343G] Compress Strings 题解2024-03-1332.CF494C Helping People 题解2024-03-2333.CF922E Birds 题解2024-03-2334.CF1628D1 Game on Sum (Easy Version) 题解2024-03-2335.P7137 [THUPC2021 初赛] 切切糕 题解2024-03-2736.CF1392H ZS Shuffles Cards 题解2024-03-2837.luogu P1543 [POI2004] SZP 题解2024-03-3038.SP64 PERMUT1 - Permutations 题解2024-04-0739.AT_s8pc_2_e 部分文字列 题解2024-04-0740.UVA1223 Editor 题解2024-04-0741.UVA10870 Recurrences 题解2024-05-0242.UVA1362 Exploring Pyramids 题解2024-05-0243.CF1141E Superhero Battle 题解2024-05-2644.SP14943 DRTREE - Dynamically-Rooted Tree 题解2024-05-2645.UVA11922 Permutation Transformer 题解2024-05-2646.UVA1674 闪电的能量 Lightning Energy Report 题解2024-05-2647.P10499 开关问题 题解2024-06-0848.P10466 邻值查找 题解2024-06-0849.P10596 BZOJ2839 集合计数 题解2024-06-2450.UVA12369 Cards 题解2024-06-2851.CF696B Puzzles 题解2024-06-2852.P10672 【MX-S1-T1】壁垒 题解2024-06-2953.P10673 【MX-S1-T2】催化剂 题解2024-06-2954.UVA11181 条件概率 Probability|Given 题解2024-06-2955.SP8177 JZPEXT - Beautiful numbers EXTREME 题解2024-07-0256.AT_tdpc_number 数 题解2024-07-0257.AT_dp_y Grid 2 题解2024-07-0358.P7690 [CEOI2002] A decorative fence 题解2024-07-0459.SP15620 POSTERIN - Postering 题解2024-07-0460.SP14887 GOODA - Good Travels 题解2024-07-1461.P2188 小Z的 k 紧凑数 题解2024-07-1462.SP8099 TABLE - Crash´s number table 题解2024-07-3063.AT_abl_e Replace Digits 题解2024-08-0564.AT_past202010_m 筆塗り 题解2024-08-0965.P8575 「DTOI-2」星之河 题解2024-08-0966.CF641E Little Artem and Time Machine 题解2024-08-0967.P10633 BZOJ2989 数列/BZOJ4170 极光 题解2024-08-1568.P10238 [yLCPC2024] F. PANDORA PARADOXXX 题解2024-08-1669.P4271 [USACO18FEB] New Barns P 题解2024-08-1670.P8512 [Ynoi Easy Round 2021] TEST_152 题解2024-08-1871.CF1514D Cut and Stick 题解2024-08-2372.P10933 创世纪 题解2024-08-2473.P10957 环路运输 题解2024-08-2474.SP10502 VIDEO - Video game combos 题解2024-08-2575.P10956 金字塔 题解2024-08-2676.CF704B Ant Man 题解2024-09-05

77.P6638 「JYLOI Round 1」常规 题解2024-09-06

78.SP1825 FTOUR2 - Free tour II 题解2024-09-2379.P10603 BZOJ4372 烁烁的游戏 题解2024-09-2780.P3364 Cool loves touli 题解2024-09-3081.CF946G Almost Increasing Array 题解2024-10-0582.AT_abc283_g [ABC283G] Partial Xor Enumeration 题解2024-10-1083.CF959F Mahmoud and Ehab and yet another xor task 题解2024-10-1084.P3794 签到题IV 题解2024-10-1685.P7984 [USACO21DEC] Tickets P 题解2024-11-0886.P5479 [BJOI2015] 隐身术 题解2024-11-0887.CF2030D QED's Favorite Permutation 题解2024-11-1088.CF633H Fibonacci-ish II 题解2024-11-1489.AT_abc129_f [ABC129F] Takahashi's Basics in Education and Learning 题解2024-12-1890.P8060 [POI2003] Sums 题解2024-12-2191.AT_keyence2019_e Connecting Cities 题解2024-12-2392.CF1534G A New Beginning 题解2024-12-2693.AT_abc237_g [ABC237G] Range Sort Query 题解2024-12-3094.CF601E A Museum Robbery 题解01-0195.P5680 [GZOI2017] 共享单车 题解01-0496.P3081 [USACO13MAR] Hill Walk G 题解01-0697.AT_abc248_h [ABC248Ex] Beautiful Subsequences 题解01-10

收起

题目传送门

前置知识

可持久化线段树 | 前缀和 & 差分

解法

进行差分，区间查询转化成前缀和相减。

先将 $\{a\}$ 升序排序。

设当前询问的区间为 $[1,r]$，在 $\{a\}$ 中找到一个最大的 $pos$ 使得 $a_{pos}\le r$，则 $[1,r]$ 中所做常规的次数为 $\sum _{i=1}^{pos}\lfloor {\displaystyle \frac{r-a_i}{k}}\rfloor$。

考虑拆开向下取整，有 $\begin{array}{rl}& \sum _{i=1}^{pos}\lfloor {\displaystyle \frac{r-a_i}{k}}\rfloor \\ & =\sum _{i=1}^{pos}{\displaystyle \frac{r-a_i-(r-a_i)modk}{k}}\\ & ={\displaystyle \frac{1}{k}}\times \sum _{i=1}^{pos}r-a_i-(r-a_i)modk\\ & ={\displaystyle \frac{1}{k}}\times (r\times pos-\sum _{i=1}^{pos}{a}_i-\sum _{i=1}^{pos}(r-a_i)modk)\end{array}$。

难点在于怎么求 $\sum _{i=1}^{pos}(r-a_i)modk$。考虑按照取模运算的性质拆开并适当补上 $+k$，有 $\begin{array}{r}(r-a_i)modk=rmodk-a_{i}modk+[rmodk<a_{i}modk]\times k\end{array}$。

以 $a_{i}modk$ 为权值建立一棵主席树然后主席树上查询 $[1,pos]$ 内满足 $a_{i}modk\in (rmodk,k)$ 即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define ll long long 
#define ull unsigned long long
#define sort stable_sort 
#define endl '\n'
ll a[100010],sum1[100010],sum2[100010];
struct PDS_SMT
{
	ll root[100010],rt_sum=0;
	struct SegmentTree
	{
		ll ls,rs,sum;
	}tree[100010<<5];
	#define lson(rt) tree[rt].ls
	#define rson(rt) tree[rt].rs
	ll build_rt()
	{
		rt_sum++;
		return rt_sum;
	}
	void update(ll pre,ll &rt,ll l,ll r,ll pos)
	{
		rt=build_rt();
		tree[rt]=tree[pre];
		tree[rt].sum++;
		if(l==r)
		{
			return;
		}
		ll mid=(l+r)/2;
		if(pos<=mid)
		{
			update(lson(pre),lson(rt),l,mid,pos);
		}
		else
		{
			update(rson(pre),rson(rt),mid+1,r,pos);
		}
	}
	ll query(ll rt,ll l,ll r,ll x,ll y)
	{
		if(x<=l&&r<=y)
		{
			return tree[rt].sum;
		}
		ll mid=(l+r)/2,ans=0;
		if(x<=mid)
		{
			ans+=query(lson(rt),l,mid,x,y);
		}
		if(y>mid)
		{
			ans+=query(rson(rt),mid+1,r,x,y);
		}
		return ans;
	}
}T;
ll ask(ll r,ll n,ll k)
{
	ll ans=0,pos=upper_bound(a+1,a+1+n,r)-a-1;
	ans+=r*pos;
	ans-=sum1[pos];
	ans-=(r%k)*pos;
	ans+=sum2[pos];
	ans-=T.query(T.root[pos],0,k-1,r%k+1,k-1)*k;
	return ans/k;
}
int main()
{
	ll type,n,m,k,mod,l,r,ans=0,i;
	cin>>type>>n>>m>>k;
	if(type==1)
	{
		cin>>mod;
	}
	for(i=1;i<=n;i++)
	{
		cin>>a[i];
	}
	sort(a+1,a+1+n);
	for(i=1;i<=n;i++)
	{
		sum1[i]=sum1[i-1]+a[i];
		sum2[i]=sum2[i-1]+a[i]%k;
		T.update(T.root[i-1],T.root[i],0,k-1,a[i]%k);
	}
	cin>>l>>r;
	ans=ask(r,n,k)-ask(l-1,n,k);
	cout<<ans<<endl;
	for(i=2;i<=m;i++)
	{
		cin>>l>>r;
		if(type==1)
		{
			l=(l+ans-1)%mod+1;
			r=(r+ans-1)%mod+1;
			if(l>r)
			{
				swap(l,r);
			}
		}
		ans=ask(r,n,k)-ask(l-1,n,k);
		cout<<ans<<endl;
	}
	return 0;
}