队列（裸题）

Queue   Aizu - ALDS1_3_B

For example, we have the following queue with the quantum of 100ms.

A(150) - B(80) - C(200) - D(200)

First, process A is handled for 100ms, then the process is moved to the end of the queue with the remaining time (50ms).

B(80) - C(200) - D(200) - A(50)

Next, process B is handled for 80ms. The process is completed with the time stamp of 180ms and removed from the queue.

C(200) - D(200) - A(50)

Your task is to write a program which simulates the round-robin scheduling .

Input

n q

name₂ time₂

...

name_n time_n

In the first line the number of processes n and the quantum q are given separated by a single space.

In the following n lines, names and times for the n processes are given.
name_i and time_i are separated by a single space.

Output

For each process, prints its name and the time the process finished in order.

Constraints

• 1 ≤ n ≤ 100000
• 1 ≤ q ≤ 1000
• 1 ≤ time_i ≤ 50000
• 1 ≤ length of name_i ≤ 10
• 1 ≤ Sum of time_i ≤ 1000000

Sample Input 1

5 100
p1 150
p2 80
p3 200
p4 350
p5 20

Sample Output 1

p2 180
p5 400
p1 450
p3 550
p4 800

Notes

Template in C

题目也是一个很裸的队列，就是支持插入和删除的操作，只不过做成了环，减少了极大的空间花费。同样数组模拟队列。做成环的环的话要注意取模问题。

 1     #include <iostream>
 2     #include <cstdio>
 3     using namespace std;
 4
 5     const int maxn = 100000 + 5;
 6
 7     struct node{
 8       char my_name[10];
 9       int my_time;
10     };
11
12     int n,q;
13     int head,tail;
14     int q_time = 0;
15     node que[maxn];
16
17     void push_back(node& k){
18       que[tail++] = k;
19       tail %= (n+5);
20     }
21
22     node pop_front(){
23       node k = que[head++];
24       head %= (n+5);
25       return k;
26     }
27
28     int main(){
29       scanf("%d%d",&n,&q);
30       head = 0;
31       tail = 0;
32       for(int i = 1;i <= n; i++){
33         node Q;
34         scanf("%s%d",Q.my_name,&Q.my_time);
35         push_back(Q);
36       }
37       while(head != tail){
38         node k = pop_front();
39         if(k.my_time - q > 0){
40           k.my_time -= q;
41           push_back(k);
42           q_time += q;
43         }
44         else{
45           q_time += k.my_time;
46           printf("%s %d\n",k.my_name,q_time);
47         }
48       }
49     }