GTY has nn 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 aiai , 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][l,r] . Because of GTY's strange hobbies, he wants there is a permutation [1..r−l+1][1..r−l+1] in [l,r][l,r]. You need to let him know if there is such a permutation or not.
InputMulti test cases (about 3) . The first line contains two integers n and m ( 1≤n,m≤10000001≤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 ithith number aiai ( 1≤ai≤n1≤ai≤n ) indicates GTY's ithith 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≤n1≤l≤r≤n ), indicating the query range.OutputFor each query, if there is a permutation [1..r−l+1][1..r−l+1] in [l,r][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
题意 给出一个数列，询问连续的从l开始到r为止的数是否刚好能够组成从1开始到r-l+1的数列
第一个判断用前缀和求和，然后就是查重
用线段树维护每一个数它上一次出现的地方，
如果这个区间里面所有的数上一次出现的最大值小于了a 那就是符合条件的了
1 #include <cstdio> 2 #include <cstring> 3 #include <queue> 4 #include <cmath> 5 #include <algorithm> 6 #include <set> 7 #include <iostream> 8 #include <map> 9 #include <stack> 10 #include <string> 11 #include <vector> 12 #define pi acos(-1.0) 13 #define eps 1e-6 14 #define fi first 15 #define se second 16 #define rtl rt<<1 17 #define rtr rt<<1|1 18 #define bug printf("****** ") 19 #define mem(a,b) memset(a,b,sizeof(a)) 20 #define name2str(x) #x 21 #define fuck(x) cout<<#x" = "<<x<<endl 22 #define f(a) a*a 23 #define sf(n) scanf("%d", &n) 24 #define sff(a,b) scanf("%d %d", &a, &b) 25 #define sfff(a,b,c) scanf("%d %d %d", &a, &b, &c) 26 #define sffff(a,b,c,d) scanf("%d %d %d %d", &a, &b, &c, &d) 27 #define pf printf 28 #define FRE(i,a,b) for(i = a; i <= b; i++) 29 #define FREE(i,a,b) for(i = a; i >= b; i--) 30 #define FRL(i,a,b) for(i = a; i < b; i++)+ 31 #define FRLL(i,a,b) for(i = a; i > b; i--) 32 #define FIN freopen("in.txt","r",stdin) 33 #define gcd(a,b) __gcd(a,b) 34 #define lowbit(x) x&-x 35 using namespace std; 36 typedef long long LL; 37 typedef unsigned long long ULL; 38 const int mod = 1e9 + 7; 39 const int maxn = 1e6 + 10; 40 const int INF = 0x3f3f3f3f; 41 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL; 42 int n, m, x, vis[maxn]; 43 LL sum[maxn]; 44 struct node { 45 int l, r, num; 46 int mid() { 47 return ( l + r ) >> 1; 48 } 49 } tree[maxn << 2]; 50 void pushup ( int rt ) { 51 tree[rt].num = max ( tree[rtl].num, tree[rtr].num ); 52 } 53 void build ( int l, int r, int rt ) { 54 tree[rt].l = l, tree[rt].r = r; 55 if ( l == r ) { 56 tree[rt].num = 0; 57 return ; 58 } 59 int m = ( l + r ) >> 1; 60 build ( l, m, rtl ); 61 build ( m + 1, r, rtr ); 62 } 63 void update ( int pos, int key, int rt ) { 64 if ( tree[rt].l == tree[rt].r ) { 65 tree[rt].num += key; 66 return ; 67 } 68 int m = tree[rt].mid(); 69 if ( pos <= m ) update ( pos, key, rtl ); 70 else update ( pos, key, rtr ); 71 pushup ( rt ); 72 } 73 int query ( int L, int R, int rt ) { 74 if ( L <= tree[rt].l && tree[rt].r <= R ) return tree[rt].num; 75 int m = tree[rt].mid(); 76 if ( L > m ) return query ( L, R, rtr ); 77 else if ( R <= m ) return query ( L, R, rtl ); 78 else return max ( query ( L, m, rtl ), query ( m + 1, R, rtr ) ); 79 } 80 int main() { 81 while ( ~sff ( n, m ) ) { 82 mem ( vis, 0 ); 83 build ( 1, n, 1 ); 84 for ( int i = 1 ; i <= n ; i++ ) { 85 sf ( x ); 86 sum[i] = sum[i - 1] + x; 87 if ( vis[x] ) update ( i, vis[x], 1 ); 88 vis[x] = i; 89 } 90 while ( m-- ) { 91 int a, b; 92 sff ( a, b ); 93 // fuck(sum[b] - sum[a - 1]),fuck(( b - a + 1 ) * ( b - a + 2 ) / 2) 94 if ( sum[b] - sum[a - 1] == ( b - a + 1 ) * ( b - a + 2 ) / 2 && query ( a, b, 1 ) < a ) printf ( "YES " ); 95 else printf ( "NO " ); 96 } 97 } 98 return 0; 99 }