Jeronimo's List(模拟+计数排序||桶排)

眉间皱痕 提交于 2019-11-27 05:08:57

 Jeronimo the bear loves numbers and he is planning to write n numbers in his notebook.

 After writing the first m numbers, Jeronimo felt that he was spending a lot of time thinking new numbers, so he wrote the next n - m missing numbers as the sum modulo 3 × 107 of the numbers in the i - m and i - m + 1 positions for m < i ≤ n

 While Jeronimo was writing, his sister Lupe arrived and asked him q questions. The i - thquestion consist of a number bi, Jeronimo has to say what would be the number in the position biif all the numbers were sorted in ascending order. Jeronimo wants to answer each question as soon as possible but he spends a lot of time counting so he ask your help.

 

Input

 The first line of the input has three integers n(3 ≤ n ≤ 3 × 107), m (3 ≤ m ≤ min(100, n)) and q(1 ≤ q ≤ 10000).

 The second line contains m numbers a1, a2, ..., am(0 ≤ ai < 3 × 107), The first m numbers that Jeronimo wrote.

 The third line contains q questions b1, b2, ..., bq(1 ≤ bi ≤ n)

 

Output

 Print q lines. The i - th line must be the answer of the i - th question made by Lupe.

 

Examples

Input
6 3 61 2 31 2 3 4 5 6
Output
123356
Input
10 4 31 2 9 101 5 10
Output
11030
开始没注意3e7开始对整个数组sort排了下序,T了,
后来二分查第x小的数,T了,又另加了vis数组记忆化标记计算过的,也T了,
愣是没想过桶排
 
计数存,俗称桶排,代码超友好:
 
 1 #include <stdio.h>
 2 #include <string.h>
 3 
 4 #define mod 30000000
 5 #define MAX 30000007
 6 
 7 int a[MAX], num[MAX];
 8 int re[MAX];
 9 
10 int main()
11 {
12     int n, i, j, pos, m, q, cur;
13     scanf("%d %d %d", &n, &m, &q);
14     memset(num, 0, sizeof(num));
15     
16     //数组里的数计数到num数组里
17     for(i=0; i<m; i++)
18     {
19         scanf("%d", &a[i]);
20         num[a[i]]++;;
21     }
22     for(; i<n; i++)
23     {
24         a[i] = (a[i-m] + a[i-m+1])%mod;
25         num[a[i]]++;
26     }
27     
28     //数组re是利用计数数组num的到的排序的后数组
29     cur = 0;
30     for(i=0;i<MAX;i++)
31     {
32         for(j=0;j<num[i];j++)
33         {
34             re[cur++] = i;
35         }
36     }
37     
38     while(q--)
39     {
40         scanf("%d", &pos);
41         printf("%d\n", re[pos-1]);
42     }
43     return 0;
44 }
 
今天写的博客超水,莫见怪
 
 
 
 
 
标签
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!