试验二

#include<stdio.h>
#include<string.h>
#define N 100
struct JCB{
    char name[10];
    int atime;
    int stime;
    int ftime;
    float ztime;
    float dcztime;
    };
    JCB a[N];
    void FCFS(int number,JCB a[N]);

void main()
{
    int number,i;

    printf("please input the number of the job: ");
    scanf("%d",&number);
    for(i=0;i<number;i++)
    {
        printf("name:");
        scanf("%s",&a[i].name);
        printf("atime:");
        scanf("%d",&a[i].atime);
        printf("stime:");
        scanf("%d",&a[i].stime);
    }

    FCFS(number,a);
}

void FCFS(int number, JCB a[N])
{
    int i,j,l,y;
    char k[N];
    for(i=0;i<number;i++) //按进程到达时间的先后排序
 {                               //如果两个进程同时到达，按在屏幕先输入的先运行
  for(j=i+1;j<number;j++)
  {
   if(a[j].atime<a[i].atime)
   {
    l=a[j].atime;
    a[j].atime=a[i].atime;
    a[i].atime=l;
    y=a[j].stime;
    a[j].stime=a[i].stime;
    a[i].stime=y;
    strcpy(k,a[j].name);
    strcpy(a[j].name,a[i].name);
    strcpy(a[i].name,k);
   }
  }
 }
for(i=0;i<number;i++)
{
    a[i].ftime=a[i].atime+a[i].stime;
    a[i].ztime=a[i].ftime-a[i].atime;
}
 for(i=0;i<number;i++)
 {
     // a[i].ftime=a[i].atime+a[i].runtime;
      printf("%5s%5d%5d\n",a[i].name,a[i].atime,a[i].stime);
    //  a[i+1].atime=a[i].ftime;
 }

}

做了，先来先服务的排序，还没完成先来先服务，不太会做。