尺取法

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.

1800: [Ahoi2009]fly 飞行棋 # Time Limit: 10 Sec Memory Limit: 64 MB Submit: 1530 Solved: 1220 [Submit][Status][Discuss]

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 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。

cf660C solution:ruler.1A 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.

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。

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.

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

不多说，直接代码。 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}

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

Jessica’s Reading Problem **Time Limit:** 1000MS **Memory Limit:** 65536K **Total Submissions:** 8787 **Accepted:** 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.