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
name₁ time₁
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。

    #include <iostream>  
    #include <cstdio>  
    using namespace std;  
      
    const int maxn = 100000 + 5;  
      
    struct node{  
      char my_name[10];  
      int my_time;  
    };  
      
    int n,q;  
    int head,tail;  
    int q_time = 0;  
    node que[maxn];  
      
    void push_back(node& k){  
      que[tail++] = k;  
      tail %= (n+5);  
    }  
      
    node pop_front(){  
      node k = que[head++];  
      head %= (n+5);  
      return k;  
    }  
      
    int main(){  
      scanf("%d%d",&n,&q);  
      head = 0;  
      tail = 0;  
      for(int i = 1;i <= n; i++){  
        node Q;  
        scanf("%s%d",Q.my_name,&Q.my_time);  
        push_back(Q);  
      }  
      while(head != tail){  
        node k = pop_front();  
        if(k.my_time - q > 0){  
          k.my_time -= q;  
          push_back(k);  
          q_time += q;  
        }  
        else{  
          q_time += k.my_time;  
          printf("%s %d\n",k.my_name,q_time);  
        }  
      }  
    }