hdu 3413 poj 3778 Single CPU, multi-tasking

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Single CPU, multi-tasking

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 557    Accepted Submission(s): 193

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

231.5 4.2 2.853.5 4.2 1.6 3.8 4.4

 

Sample Output

Case 1:3.508.507.30Case 2:14.1017.107.6015.9017.50

 

Source

2010 National Programming Invitational Contest Host by ZSTU

题意：就是输出每个任务完成的时间！任务在完成之前是同时进行的，意思就是在第一个任务进行一毫秒后就只能进行第二个任务一毫秒，以此类推！

#include <cstdio>#include <cmath>#include <cstring>#include <string>#include <cstdlib>#include <climits>#include <ctype.h>#include <queue>#include <stack>#include <vector>#include <deque>#include <set>#include <map>#include <iostream>#include <algorithm>using namespace std;#define PI acos(-1.0)#define INF 0xffffffdouble MIN(double a,double b){if( a < b)return a;return b;}int main(){int i,j,k = 0,n,t;double min,time,temp;double use[1047],need[1047];while(~scanf("%d",&t)){while(t--){k++;memset(use,0,sizeof(use));memset(need,0,sizeof(need));time = 0;scanf("%d",&n);for( i = 0 ; i < n ; i++){scanf("%lf",&need[i]);}for( i = 0 ; i < n ; i++){min = 10000047;for(j = 0 ; j < n ; j++)//每次找最小的{if(need[j] == 0)continue;temp =(int)need[j];if(temp == need[j])//如果最小的数是整数，要减1,{//避免在min>1时直接出现need[j]-(int)min==0的情况temp--;} min = MIN(temp,min);}for(j = 0 ; j < n ; j++)//优化，每次直接减去最小的数{if(need[j] == 0)continue;need[j]-=(int)min;//但是不应该直接出现need[j]-(int)min==0的情况time+=(int)min;}for(j = 0 ; j < n; j++){if(need[j] == 0)continue;if(need[j] <= 1){time+=need[j];use[j]=time;need[j] = 0;}else{need[j] -= 1;time +=1;}}}printf("Case %d:\n",k);for(i = 0 ; i < n ; i++){printf("%.2lf\n",use[i]);}}}}