双指针 on 111qqz's blog

leetcode 209. Minimum Size Subarray Sum (尺取法)

hust.111qqz@gmail.com (111qqz) — Thu, 13 Apr 2017 13:50:02 +0000

Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn’t one, return 0 instead.

leetcode 75. Sort Colors

hust.111qqz@gmail.com (111qqz) — Thu, 13 Apr 2017 12:02:02 +0000

Given an array with n objects colored red, white or blue, sort them so that objects of the same color are adjacent, with the colors in the order red, white and blue.

leetcode 11. Container With Most Water (two pointer)

hust.111qqz@gmail.com (111qqz) — Thu, 13 Apr 2017 10:13:01 +0000

Given n non-negative integers a1, a2, …, an, where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.

leetcode 16. 3Sum Closest (k-sum问题，two pointer)

hust.111qqz@gmail.com (111qqz) — Thu, 13 Apr 2017 09:45:45 +0000

Given an array S of n integers, find three integers in S such that the sum is closest to a given number, target. Return the sum of the three integers. You may assume that each input would have exactly one solution.

leetcode 18. 4Sum (k-sum问题，two pointer)

hust.111qqz@gmail.com (111qqz) — Thu, 13 Apr 2017 09:34:15 +0000

Given an array S of n integers, are there elements a, b, c, and d in S such that a + b + c + d = target? Find all unique quadruplets in the array which gives the sum of target.

leetcode 15. 3Sum (k-sum问题，two pointer)

hust.111qqz@gmail.com (111qqz) — Thu, 13 Apr 2017 08:21:52 +0000

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.

BZOJ 1800: [Ahoi2009]fly 飞行棋（尺取+数学）

hust.111qqz@gmail.com (111qqz) — Thu, 26 Jan 2017 15:42:07 +0000

1800: [Ahoi2009]fly 飞行棋
#

Time Limit: 10 Sec Memory Limit: 64 MB Submit: 1530 Solved: 1220 [Submit][Status][Discuss]

codeforces 220 E. Little Elephant and Inversions (树状数组+尺取)

hust.111qqz@gmail.com (111qqz) — Thu, 15 Sep 2016 08:13:56 +0000

题目链接

题意：

how many pairs of integers l and r are there, such that 1 ≤ l < r ≤ n and sequence b = _a_1_a_2… a__l__a__r__a__r + 1… a__n has no more than k inversions.

hdu 4123 Bob’s Race (树的直径+尺取+rmq)(珍爱生命，远离log)

hust.111qqz@gmail.com (111qqz) — Sun, 17 Jul 2016 16:21:03 +0000

hdu 4123 题目链接

题意：一棵树，定义 d[i] 为点 i 到树上某点的最大距离。给出若干查询，每个查询一个 x，问最多能有多少点满足这些点中，最大的 d 与最小的 d 的差小于等于 x。要求这些点的编号必须是连续的。

codeforces 660 C. Hard Process (ruler)

hust.111qqz@gmail.com (111qqz) — Wed, 08 Jun 2016 16:17:00 +0000

cf660C

solution:ruler.1A

代码实现

 1/* ***********************************************
 2Author :111qqz
 3Created Time :2016年06月08日 星期三 23时43分18秒
 4File Name :code/cf/problem/660C.cpp
 5************************************************ */
 6
 7#include <cstdio>
 8#include <cstring>
 9#include <iostream>
 10#include <algorithm>
 11#include <vector>
 12#include <queue>
 13#include <set>
 14#include <map>
 15#include <string>
 16#include <cmath>
 17#include <cstdlib>
 18#include <ctime>
 19#define fst first
 20#define sec second
 21#define lson l,m,rt<<1
 22#define rson m+1,r,rt<<1|1
 23#define ms(a,x) memset(a,x,sizeof(a))
 24typedef long long LL;
 25#define pi pair < int ,int >
 26#define MP make_pair
 27
 28using namespace std;
 29const double eps = 1E-8;
 30const int dx4[4]={1,0,0,-1};
 31const int dy4[4]={0,-1,1,0};
 32const int inf = 0x3f3f3f3f;
 33const int N=3E5+7;
 34int n,k;
 35int sum[N],a[N];
 36
 37void ruler()
 38{
 39 int head = 1;
 40 int tail = 1;
 41 int l,r;
 42 int res = -1;
 43 int cnt = 0 ;
 44
 45 while (tail<=n)
 46 {
 47	while (a[tail]==1) tail++;
 48//	cout<<"head:"<<head<<" tail:"<<tail<<endl;
 49	if (a[tail]==0&&tail<=n) cnt++;
 50
 51	while (sum[tail]-sum[head-1]<=k&&tail<=n) tail++;
 52//	cout<<"head:"<<head<<"tail:"<<tail<<endl;
 53	 if (tail-head>res)
 54	 {
 55		res = tail-head;
 56	//	cout<<"res:"<<res<<endl;
 57		l = head;
 58		r = tail-1;
 59	 }
 60
 61	while (head<=tail&&sum[tail]-sum[head-1]>k) head++;
 62//	cout<<"head::"<<head<<" tail:"<<tail<<endl;
 63	if (tail<=n&&tail-head+1>res)
 64	{
 65	 res = tail-head+1;
 66	 l = head;
 67	 r = tail;
 68	}
 69
 70
 71 }
 72
 73
 74 for ( int i = l ; i <= r ; i++) a[i] = 1;
 75
 76 cout<<res<<endl;
 77 for ( int i = 1 ; i <= n ; i++) cout<<a[i]<<" ";
 78}
 79int main()
 80{
 81	#ifndef ONLINE_JUDGE
 82	freopen("code/in.txt","r",stdin);
 83 #endif
 84
 85	cin>>n>>k;
 86
 87	sum[0] = 0;
 88	for ( int i = 1; i <= n ; i++)
 89	{
 90	 scanf("%d",&a[i]);
 91	 sum[i] = sum[i-1] +(1-a[i]);
 92	}
 93
 94	ruler();
 95
 96
 97 #ifndef ONLINE_JUDGE
 98 fclose(stdin);
 99 #endif
100 return 0;
101}

hdu 3530 Subsequence (尺取+rmq)

hust.111qqz@gmail.com (111qqz) — Thu, 19 May 2016 09:44:36 +0000

hdu 3530题目链接

题意：给出n个数，m,k，问最大的j-i+1，使得【i,j】间的最大值与最小值的差属于[m,k] 思路：rmq+尺取。 2A.

codeforces 279 B books

hust.111qqz@gmail.com (111qqz) — Tue, 15 Dec 2015 12:23:32 +0000

http://codeforces.com/problemset/problem/279/B 题意：给定一个序列，问一段连续的序列的和小于等于t的最长的序列的长度。思路：尺取法。三个月前学习的了。

codeforces #321 div 2 B. Kefa and Company(尺取法)

hust.111qqz@gmail.com (111qqz) — Fri, 25 Sep 2015 14:46:00 +0000

B. Kefa and Company

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Kefa wants to celebrate his first big salary by going to restaurant. However, he needs company.

poj 2739 Sum of Consecutive Prime Numbers (尺取法)

hust.111qqz@gmail.com (111qqz) — Fri, 25 Sep 2015 05:27:00 +0000

一开始迷之wa… 先找出素数下标的上界就可以A… 然后纠结了２０分钟．．．然后发现是预处理的素数少了一个素数．．我预处理是处理到＜１０００５的素数．．．最大数１００００，而超过１００００的第一个素数是１０００７这样判断终止条件就会死循环… sad

poj 2100 Graveyard Design (two pointers ，尺取法)

hust.111qqz@gmail.com (111qqz) — Thu, 24 Sep 2015 17:29:00 +0000

不多说，直接代码。

代码实现

 1/*************************************************************************
 2	> File Name: code/poj/2100.cpp
 3	> Author: 111qqz
 4	> Email: rkz2013@126.com
 5	> Created Time: 2015年09月25日 星期五 00时42分49秒
 6 ************************************************************************/
 7
 8#include<iostream>
 9#include<iomanip>
10#include<cstdio>
11#include<algorithm>
12#include<cmath>
13#include<cstring>
14#include<string>
15#include<map>
16#include<set>
17#include<queue>
18#include<vector>
19#include<stack>
20#include<cctype>
21#define y1 hust111qqz
22#define yn hez111qqz
23#define j1 cute111qqz
24#define ms(a,x) memset(a,x,sizeof(a))
25#define lr dying111qqz
26using namespace std;
27#define For(i, n) for (int i=0;i<int(n);++i)
28typedef long long LL;
29typedef double DB;
30const int inf = 0x3f3f3f3f;
31LL n;
32LL maxn;
33vector<LL>ans;
34
35
36void solve()
37{
38 LL head = 1,tail = 1;
39 LL sum = 0 ;
40
41 while (tail<=maxn)
42 {
43	sum = sum + tail*tail;
44
45	if (sum>=n)
46	{
47	 while (sum>n)//主要是while，因为可能要减掉多个才能小于n
48	 {
49		sum -= head*head;
50		head++;
51	 }
52	 if (sum==n)
53	 {//因为要先输出答案个数...所以必须存起来延迟输出...
54		ans.push_back(head);
55		ans.push_back(tail);
56	 }
57	}
58	tail++;
59 }
60 LL sz = ans.size();
61 printf("%lld\n",sz/2);
62 for (LL i = 0 ; i<ans.size() ; i = i +2)
63 {
64	printf("%lld",ans[i+1]-ans[i]+1);
65	for ( LL j = ans[i] ; j <=ans[i+1] ; j++) printf(" %lld",j);
66	printf("\n");
67 }
68
69}
70int main()
71{
72 #ifndef ONLINE_JUDGE
73 freopen("in.txt","r",stdin);
74 #endif
75 scanf("%lld",&n);
76 maxn = ceil(sqrt(n));
77 solve();
78
79 #ifndef ONLINE_JUDGE
80 fclose(stdin);
81 #endif
82	return 0;
83}

poj 2566 Bound Found (前缀和，尺取法（two pointer）)

hust.111qqz@gmail.com (111qqz) — Thu, 24 Sep 2015 16:26:00 +0000

题意　：给定一个长度为n的区间．然后给k次询问，每次一个数t，求一个区间[l,r]使得这个区间和的绝对值最接近t

poj 3320 Jessica's Reading Problem (尺取法)

hust.111qqz@gmail.com (111qqz) — Thu, 24 Sep 2015 13:37:00 +0000

Jessica’s Reading Problem

Time Limit	Memory Limit	Total Submissions	Accepted
1000MS	65536K	8787	2824

Description

Jessica’s a very lovely girl wooed by lots of boys. Recently she has a problem. The final exam is coming, yet she has spent little time on it. If she wants to pass it, she has to master all ideas included in a very thick text book. The author of that text book, like other authors, is extremely fussy about the ideas, thus some ideas are covered more than once. Jessica think if she managed to read each idea at least once, she can pass the exam. She decides to read only one contiguous part of the book which contains all ideas covered by the entire book. And of course, the sub-book should be as thin as possible.

迟取法

hust.111qqz@gmail.com (111qqz) — Thu, 24 Sep 2015 11:02:00 +0000