描述

Cinema in Akiba (CIA) is a small but very popular cinema in Akihabara.
Every night the cinema is full of people. The layout of CIA is very
interesting, as there is only one row so that every audience can enjoy
the wonderful movies without any annoyance by other audiences sitting
in front of him/her.

The ticket for CIA is strange, too. There are n seats in CIA and they
are numbered from 1 to n in order. Apparently, n tickets will be sold
everyday. When buying a ticket, if there are k tickets left, your
ticket number will be an integer i (1 ≤ i ≤ k), and you should choose
the ith empty seat (not occupied by others) and sit down for the film.

On November, 11th, n geeks go to CIA to celebrate their anual
festival. The ticket number of the ith geek is ai. Can you help them
find out their seat numbers?

Input

The input contains multiple test cases. Process to end of file. The
first line of each case is an integer n (1 ≤ n ≤ 50000), the number of
geeks as well as the number of seats in CIA. Then follows a line
containing n integers a1, a2, …, an (1 ≤ ai ≤ n - i + 1), as
described above. The third line is an integer m (1 ≤ m ≤ 3000), the
number of queries, and the next line is m integers, q1, q2, …, qm (1
≤ qi ≤ n), each represents the geek’s number and you should help him
find his seat.

Output

For each test case, print m integers in a line, seperated by one
space. The ith integer is the seat number of the qith geek.

Sample Input

Sample Output

1 2 3
4 5 3 1 2

思路

先说题意，电影院里面有n个座位，作为的序号是从1~n，现在有n个人要坐座位,每个票上面有一个数字a[i],这个数字代表这个人应该坐在剩下的空位中的第几个，按照题目给出的顺序依次坐，最后有m组询问，问谁坐在哪个位置。

我们用线段树来位数剩下的座位的数量，初始化的时候线段树每个节点为1，更新的时候查看左儿子有没有位置，然后再看右儿子，更新即可

代码

#include <cstdio>
#include <cstring>
#include <cctype>
#include <stdlib.h>
#include <string>
#include <map>
#include <iostream>
#include <sstream>
#include <set>
#include <stack>
#include <cmath>
#include <queue>
#include <vector>
#include <algorithm>
#include <list>
using namespace std;
#define mem(a, b) memset(a, b, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define inf 0x3f3f3f3f
typedef long long ll;
const int N = 50000 + 20;
int sum[N << 2], ans[N];
void pushup(int rt)
{sum[rt] = sum[rt << 1] + sum[rt << 1 | 1];
}
void build(int l, int r, int rt)
{if (l == r){sum[rt] = 1;return;}int m = (l + r) >> 1;build(lson);build(rson);pushup(rt);
}
int update(int l, int r, int rt, int x)
{if (l == r){sum[rt] = 0;return l;}int ret, m = (l + r) >> 1;if (x <= sum[rt << 1])ret = update(lson, x);elseret = update(rson, x - sum[rt << 1]);pushup(rt);return ret;
}
int main()
{//freopen("in.txt", "r", stdin);int n, x, m;while (~scanf("%d", &n)){mem(ans, 0);build(1, n, 1);for (int i = 1; i <= n; i++){scanf("%d", &x);ans[i] = update(1, n, 1, x);}scanf("%d", &m);for (int i = 1; i <= m; i++){scanf("%d", &x);if (i == 1)printf("%d", ans[x]);elseprintf(" %d", ans[x]);}puts("");}return 0;
}

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ZOJ3635 Cinema in Akiba(线段树)

描述

Input

Output

Sample Input

Sample Output

思路

代码

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