Cinema in Akiba
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 theith 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
3 1 1 1 3 1 2 3 5 2 3 3 2 1 5 2 3 4 5 1
Sample Output
1 2 3 4 5 3 1 2
思路比较清晰。
树状数组+二分
具体见注释
1 /************************************************************************* 2 > File Name: code/zoj/3635.cpp 3 > Author: 111qqz 4 > Email: rkz2013@126.com 5 > Created Time: 2015年10月22日 星期四 10时03分36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 22 #define yn hez111qqz 23 #define j1 cute111qqz 24 #define ms(a,x) memset(a,x,sizeof(a)) 25 using namespace std; 26 const int dx4[4]={1,0,0,-1}; 27 const int dy4[4]={0,-1,1,0}; 28 typedef long long LL; 29 typedef double DB; 30 const int inf = 0x3f3f3f3f; 31 const int N=5E5+7; 32 int t[N],a[N]; 33 int ans[N],k; 34 int n,m; 35 36 37 int lowbit( int x) 38 { 39 return x&(-x); 40 } 41 void update ( int x,int delta) 42 { 43 for ( int i = x ; i <= n ; i = i + lowbit(i)) 44 { 45 t[i] = t[i] + delta; 46 } 47 } 48 49 int sum( int x) 50 { 51 int res = 0 ; 52 for ( int i = x; i >= 1 ; i = i - lowbit(i)) 53 { 54 res = res + t[i]; 55 } 56 return res; 57 } 58 59 int bin_search(int l,int r) 60 { 61 while (l<r) 62 { 63 int mid = (l+r)>>1; 64 if (sum(mid)<k) 65 l = mid + 1; 66 else r = mid ; 67 } 68 return r; 69 } 70 int main() 71 { 72 #ifndef ONLINE_JUDGE 73 freopen("in.txt","r",stdin); 74 #endif 75 76 while (scanf("%d",&n)!=EOF) 77 { 78 79 ms(t,0); 80 ms(ans,0); 81 ms(a,0); 82 for ( int i = 1 ; i <= n ; i++) update(i,1); //1表示没有被占，初始所有位置都没有被占。 83 84 for ( int i = 1 ; i <= n ; i++) 85 { 86 87 scanf("%d",&k); //每次要占第k个没有被占的位置，由于被占的位置被设置成0,所以就是找第k大的。 88 89 int posi = bin_search(1,n); 90 ans[i] =posi; 91 update(posi,-1); //将被占的位置设置成0 92 93 } 94 95 96 97 98 scanf("%d",&m); 99 int x; 100 for ( int i = 1 ; i < m ; i++) 101 { 102 scanf("%d",&x); 103 printf("%d ",ans[x]); 104 } 105 scanf("%d",&x); 106 printf("%d ",ans[x]); 107 108 } 109 110 111 #ifndef ONLINE_JUDGE 112 fclose(stdin); 113 #endif 114 return 0; 115 }