hdu題3413—基礎算法（模擬題）

Single CPU, multi-tasking

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 582 Accepted Submission(s): 204

Problem Description

Tuntun is a big fan of Apple Inc. who owns almost all kinds of products the company published. Fascinated by the amazing user experience and fabulous user interface, he spent every nickel of his pocket money on his iMac and iPhone. A few days ago, Apple released their latest iPhone OS 4.0. Tuntun noticed that the most significant new feature of iPhone OS 4.0 is multi-tasking support. He was curious about why the same device with a single core CPU can support multi-tasking under the new operating system. With his clever head, he found out a simple solution. The OS doesn’t have to let the CPU run several tasks exactly at the same time. What the OS should do is just to let the user feel that several tasks are running at the same time. In order to do that, the OS assigns the CPU to the tasks in turn. When the acts of reassigning a CPU from one task to another occur frequently enough, the illusion of parallelism is achieved. Let’s suppose that the OS makes each task run on the CPU for one millisecond every time in turn, and when a task is finished, the OS assigns the CPU to another task immediately. Now if there are 3 tasks which respectively need 1.5, 4.2 and 2.8 millisecond to complete, then the whole process is as follows:

At 0th millisecond, task 1 gets the CPU. After running for 1 millisecond, it still needs 0.5 milliseconds to complete.

At 1st millisecond, task 2 gets the CPU. After running for 1 millisecond, it still needs 3.2 milliseconds to complete.

At 2st millisecond, task 3 gets the CPU. After running for 1 millisecond, it still needs 1.8 milliseconds to complete.

At 3rd millisecond, task 1 comes back to CPU again. After 0.5 millisecond of running, it is finished and will never need the CPU.

At 3.5 millisecond, task 2 gets the CPU again. After running for 1 millisecond, it still needs 2.2 milliseconds to complete.

At 4.5 millisecond, it’s time for task 3 to run. After 1 millisecond, it still needs 0.8 milliseconds to complete.

At 5.5 millisecond, it’s time for task 2 to run. After 1 millisecond, it still needs 1.2 milliseconds to complete.

At 6.5 millisecond, time for task 3. It needs 0.8 millisecond to complete, so task 3 is finished at 7.3 milliseconds.

At 7.3 millisecond, task 2 takes the CPU and keeps running until it is finished.
At 8.5 millisecond, all tasks are finished.

Tuntun decided to make a simple iPhone multi-tasking OS himself, but at first, he needs to know the finishing time of every task. Can you help him?

Input

The first line contains only one integer T indicates the number of test cases.
The following 2×T lines represent T test cases. The first line of each test case is a integer N (0<N <= 100) which represents the number of tasks, and the second line contains N real numbers indicating the time needed for each task. The time is in milliseconds, greater than 0 and less than 10000000.

Output

For each test case, first output “Case N:”, N is the case No. starting form 1. Then print N lines each representing a task’s finishing time, in the order correspondent to the order of tasks in the input. The results should be rounded to 2 digits after decimal point, and you must keep 2 digits after the decimal point.

Sample Input

2 3 1.5 4.2 2.8 5 3.5 4.2 1.6 3.8 4.4

Sample Output

Case 1: 3.50 8.50 7.30 Case 2: 14.10 17.10 7.60 15.90 17.50

想法：我開始用隊列去模擬這個問題，超時了，所以又想了另外一種思路去做。具體操作如下

<span style="font-family:Arial;">#include<stdio.h>
#include<math.h>
#define MAX 200
int main(){
	int t;
	scanf("%d",&t);
	int i;
	for(i=0;i<t;i++){
		int n;
		double reserve[MAX];
		double marks=0;
		double cost[MAX];
		scanf("%d",&n);
		int j;
		for(j=0;j<n;j++){
			scanf("%lf",&reserve[j]);
		}
		int k;
		/*計算某個任務總共完成需要的時間，每個任務所在的序列的行相加就是此任務輸入的時間
		此後做其他序列上相對此任務的位置的判定，不同的位置所花費的時間是不同的，總共有n個
		任務，則需要分別計算這n個任務*/
		for(j=0;j<n;j++){
			cost[j]=reserve[j];
			for(k=0;k<n;k++){
				if(k!=j){
                    /*當某個位置時間花銷少於目標位置的情況*/
					if(floor(reserve[k]+0.999)<floor(reserve[j]+0.999))//floor(double)函數取得是某個數的下限，即不大於某個數
						cost[j]=cost[j]+reserve[k];
                    /*當某個位置的時間花銷和目標位置在一輪的時間片上的情況*/
					else if(floor(reserve[k]+0.999)==floor(reserve[j]+0.999)){
						if(k<j)
							cost[j]=cost[j]+reserve[k];
						else
							cost[j]=cost[j]+(floor(reserve[k]+0.999)-1);
					}
					/*當某個位置的時間花銷多於目標位置的情況*/
					else if(floor(reserve[k]+0.999)>floor(reserve[j]+0.999)){
						if(k>j)
							cost[j]=cost[j]+(floor(reserve[j]+0.999)-1);
						else
							cost[j]=cost[j]+floor(reserve[j]+0.999);
					}
				}
			}
		}
		printf("Case %d:\n",i+1);
		for(j=0;j<n;j++){
			printf("%.2lf\n",cost[j]);
		}
	}
	return 0;
}</span>