【正文】 Webb, the head of Fifa39。t appealed against the disciplinary action your employer has taken against you. However, if you win your case, the tribunal may reduce any pensation awarded to you as a result of your failure to appeal. Remember that in most cases you must make an application to an employment tribunal within three months of the date when the event you are plaining about happened. If your application is received after this time limit, the tribunal will not usually accept it. If you are worried about how the time limits apply to you, take advice from one of the anisations listed under Further help. Employment tribunals are less formal than some other courts, but it is still a legal process and you will need to give evidence under an oath or affirmation. Most people find making a claim to an employment tribunal challenging. If you are thinking about making a claim to an employment tribunal, you should get help straight away from one of the anisations listed under Further help. If you are being represented by a solicitor at the tribunal, they may ask you to sign an agreement where you pay their fee out of your pensation if you win the case. This is known as a damagesbased agreement. In England and Wales, your solicitor can39。 g an employment tribunal claim Employment tribunals sort out disagreements between employers and employees. You may need to make a claim to an employment tribunal if: ? you don39。 這樣的話，不僅 提高了程序的可讀性和可操作性， 而且還提高了 CPU的利用率和內(nèi)存的利用率，因為程序的運行是局部性的，這種思想對于段頁式存儲管理系統(tǒng)尤為重要。主要體會如下： Vc++編譯程序編寫銀行家算法，進一步理 解到 通過 銀行家算法避免死鎖的思想 ,同時也理解了 系統(tǒng) 死鎖產(chǎn)生的原因及條件。在確定安全序列的過程中，要檢測所有進程的 Finish[i]的值，每次循環(huán)檢測完后要重復從第一個進程開始。 return 1。 return 0。i++) //輸出安全序列 cout進程 PSequence[i] 。 for(i=0。 //試探下一個進程 } }// if(l==m)//都試探完畢 { cout系統(tǒng)安全 !endl。 //模擬資源分配序列生成 i=1。k++)//進程 i 正常運行后，釋放其占有的資源 { Work[k]+=Allocation[i][k]。 for(k=0。j++)// 找到一個滿足 Finish[i]=false 且Need[i][j]=Work[j]的進程 { if(Need[i][j]Work[j])//由于部分資源得不到滿足，進程 i 無法運行 { break。 } else //對于未運行的進程，進行如下處理 {/// for(j=0。im。i++) //掃描所有進程 ， 預設所有進程不能運行 { Finish[i]=False。 //工作分配初始化為系統(tǒng)可用資源 for(i=0。in。 //int Work[MAXn]。 else cout請輸入 1 或 2!。//回收該進程所分配的資源 coutendl。jn。 coutFinish[Sequence[i]] 。j++) coutAllocation[Sequence[i]][j]+Work[j] 。 for(j=0。j++) coutAllocation[Sequence[i]][j] 。 for(j=0。j++) coutNeed[Sequence[i]][j] 。 for(j=0。j++) coutWork[j] 。 for(j=0。im。 cout 進 程 號WorkNeedAllocationWork+Allocation Finishendl。jn。//沒有足夠的資源分配給該進程 } }//if(safealg()0) else { cout同意分配請求 !endl。im。 Need[i][j]=Need[i][j]+Request[i][j]。j++) //把資源恢復成分配之前的狀態(tài) { Available[j]=Available[j]+Request[i][j]。 for (j=0。//所得資源增加 Need[i][j]=Need[i][j]Request[i][j]。j++)//資源申請得到允許時，變換各個資源數(shù) { Available[j]=Available[j]Request[i][j]。 } } for(j=0。 } if(Request[i][j]Available[j]) { //資源申請數(shù)目大于可利用數(shù)，無法分配，得等待 cout當前系統(tǒng)可用資源不夠，請等待 !endl。j++) { if(Request[i][j]Need[i][j]) { cout申請的資源大于它需要的資源數(shù) ,請重新輸入 !\n。 //**********銀行家算法進行檢查 *************// for(j=0。jn。//重新輸入進程號 } cout請輸入進程申請的資源