hdu5172---GTY's gay friends

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 5 2 1 3 4 5 2 3 1 1 3 1 1 2 2 4 8 1 5 3 2 1 1 1 1 1 1 2

Sample Output

YES NO YES YES YES YES NO

Source
BestCoder Round #29

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如果一个区间里是一个[1, … , r - l + 1]的排列，那么首先，区间和是
x * (x + 1) / 2， –>前缀和处理
其次每一个数都不同，预处理每一个数上一次出现的位置pre，然后求出区间里pre的最大值，如果最大的pre小于左端点且区间和是 x * (x + 1) / 2，那么输出YES，否则NO，这里可以用线段树解决
注意判断时，最好先判断区间和，如果这里已经不满足了，就别去查询线段树了，否则容易造成TLE

/*************************************************************************    > File Name: hdu5172.cpp    > Author: ALex    > Mail: [e-mail]     > Created Time: 2015年02月14日 星期六 12时49分13秒 ************************************************************************/#include <map>#include <set>#include <queue>#include <stack>#include <vector>#include <cmath>#include <cstdio>#include <cstdlib>#include <cstring>#include <iostream>#include <algorithm>using namespace std;const double pi = acos(-1);const int inf = 0x3f3f3f3f;const double eps = 1e-15;typedef long long LL;typedef pair <int, int> PLL;const int N = 1000010;int pre[N];LL sum[N];int _hash[N];struct node{    int l, r;    int val;}tree[N << 2];void build (int p, int l, int r){    tree[p].l = l;    tree[p].r = r;    if (l == r)    {        tree[p].val = pre[l];        return;    }    int mid = (l + r) >> 1;    build (p << 1, l, mid);    build (p << 1 | 1, mid + 1, r);    tree[p].val = max (tree[p << 1].val, tree[p << 1 | 1].val);}int query (int p, int l, int r){    if (tree[p].l == l && tree[p].r == r)    {        return tree[p].val;    }    int mid = (tree[p].l + tree[p].r) >> 1;    if (r <= mid)    {        return query (p << 1, l, r);    }    else if (l > mid)    {        return query (p << 1 | 1, l, r);    }    else    {        return max (query (p << 1, l, mid), query (p << 1 | 1, mid + 1, r));    }}int main (){    int n, m;    while (~scanf("%d%d", &n, &m))    {        memset (_hash, -1, sizeof(_hash));        memset (sum, 0, sizeof(sum));        int x, l, r;        for (int i = 1; i <= n; ++i)        {            scanf("%d", &x);            pre[i] = _hash[x];            _hash[x] = i;            sum[i] = sum[i - 1] + (LL)x;        }        build (1, 1, n);        while (m--)        {            scanf("%d%d", &l, &r);            LL len = (LL)(r - l + 1);            len = (len + 1) * len / 2;            if (len != sum[r] - sum[l - 1])            {                printf("NO\n");                continue;            }            int res = query (1, l, r);            if (res < l)            {                printf("YES\n");            }            else            {                printf("NO\n");            }        }    }    return 0;}