HDU5172

1.题目描述：

GTY's gay friends

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1687    Accepted Submission(s): 429

Problem Description

GTY has n gay friends. To manage them conveniently, every morning he ordered all his gay friends to stand in a line. Every gay friend has a characteristic value ai , to express how manly or how girlish he is. You, as GTY's assistant, have to answer GTY's queries. In each of GTY's queries, GTY will give you a range [l,r] . Because of GTY's strange hobbies, he wants there is a permutation [1..r−l+1] in [l,r]. You need to let him know if there is such a permutation or not.

 

Input

Multi test cases (about 3) . The first line contains two integers n and m ( 1≤n,m≤1000000 ), indicating the number of GTY's gay friends and the number of GTY's queries. the second line contains n numbers seperated by spaces. The ith number ai ( 1≤ai≤n ) indicates GTY's ith gay friend's characteristic value. The next m lines describe GTY's queries. In each line there are two numbers l and r seperated by spaces ( 1≤l≤r≤n ), indicating the query range.

 

Output

For each query, if there is a permutation [1..r−l+1] in [l,r], print 'YES', else print 'NO'.

 

Sample Input

8 52 1 3 4 5 2 3 11 31 12 24 81 53 21 1 11 11 2

 

Sample Output

YESNOYESYESYESYESNO

 

Source

BestCoder Round #29

 

Recommend

hujie

 

2.题意概述：

给出一个数组a[n]（1<=a[i]<=n）,可能会有重复，然后m组询问 每次询问两个数：l,r 在区间[l,r]内是否构成一个1,2,..,r-l+1的排列；

3.解题思路：

对于[1..n][1..n][1..n]中的每一个数随机一个64位无符号整型作为它的hash值,一个集合的hash值为元素的异或和,预处理[1..n]的排列的hash和原序列的前缀hash异或和,就可以做到线性预处理,O(1)回答询问

4.AC代码：

#include <bits/stdc++.h>#define INF 0x7fffffff#define maxn 1001000#define eps 1e-6#define pi acos(-1.0)#define e 2.718281828459#define mod (int)1e9 + 7;using namespace std;typedef long long ll;int n, m;ll sum[maxn], hs[maxn], val[maxn];inline ll randval(){  ll one = 1;  ll RAND = (rand() + rand()) * (one << 44) + (rand() + rand()) * (one << 33) + (rand() + rand()) * (one << 22) + (rand() + rand());  return RAND;}int main(){#ifndef ONLINE_JUDGE  freopen("in.txt", "r", stdin);  freopen("out.txt", "w", stdout);  long _begin_time = clock();#endif  srand(time(NULL));  hs[0] = 0;  for (int i = 1; i < maxn; i++)  {    val[i] = randval();    hs[i] = hs[i - 1] ^ val[i];  }  while (scanf("%d%d", &n, &m) != EOF)  {    sum[0] = 0;    for (int i = 1; i <= n; i++)    {      int x;      scanf("%d", &x);      sum[i] = sum[i - 1] ^ val[x];    }    for (int i = 0; i < m; i++)    {      int l, r;      scanf("%d%d", &l, &r);      if (hs[r - l + 1] == (sum[r] ^ sum[l - 1]))        puts("YES");      else        puts("NO");    }  }#ifndef ONLINE_JUDGE  long _end_time = clock();  printf("time = %ld ms\n", _end_time - _begin_time);#endif  return 0;}