I can’t put up with the ______ lecture any longer, so I left my seat and began to think about the coming vacation. A. boringB. worriedC. boredD. restless
I can’t put up with the ______ lecture any longer, so I left my seat and began to think about the coming vacation.
A. boring
B. worried
C. bored
D. restless
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Lodger: I’m terribly sorry that I broke you precious vase. I’ll pay for it.Landlady:____. A.Can’t complainB.Take careC.Relax yourselfD.Never mind
1．I ______they ______tomorrow．A． think; won’ t comeB． don- t think ; comeC． don't think; will comeD． think; come
阅读下列C程序和程序说明，将应填入(n)处的字句写在对应栏内。【说明】下面是一个用C编写的快速排序算法。为了避免最坏情况，取基准记录pivot时，采用从left、right和mid=[(left+right)/2]中取中间值，并交换到right位置的办法。数组a存放待排序的一组记录，数据类型为T，left和right是待排序子区间的最左端点和最右端点。void quicksort (int a[], int left, int right) {int temp;if (left＜right) {hat pivot = median3 (a, left, right); //三者取中子程序int i = left, j = right-1;for(;;){while (i ＜j a[i] ＜ pivot) i++;while (i ＜j pivot ＜ a[j]) j--;if(i＜j){temp = a[i]; a[j] = a[i]; a[i] = temp;i++; j--;}else break;}if (a[i] ＞ pivot){temp = a[i]; a[i] = a[right]; a[right] = temp;}quicksort( (1) ); //递归排序左子区间quieksort(a,i+1 ,right); //递归排序右子区间}}void median3 (int a[], int left, int right){ int mid=(2);int k = left;if(a[mid] ＜ a[k])k = mid;if(a[high] ＜ a[k]) k = high; //选最小记录int temp = a[k]; a[k] = a[left]; a[left] = temp; //最小者交换到 leftif(a[mid] ＜ a[right]){temp=a[mid]; a[mid]=a[right]; a[right]=temp;}}消去第二个递归调用 quicksort (a,i+1,right)。 采用循环的办法:void quicksort (int a[], int left, int right) {int temp; int i,j;(3) {int pivot = median3(a, left, right); //三者取中子程序i = left; j = righi-1;for (;; ){while (i＜j a[i] ＜ pivot)i++;while (i＜j pivot ＜a[j]) j--;if(i ＜j) {temp = a[i]; a[j]; = a[i]; a[i]=temp;i++; j--;}else break;}if(a[i]＞pivot){(4);a[i]=pivot;}quicksoft ((5)); //递归排序左子区间left = i+1;}}
YouwanttouseteCoherenceJavaAPIstodirectlycachedPOJOs.Considerthissnippetofcode：NamedCachecache-CacheFactory.getCache（mycache）；cache.put（newInteger（I）fhello）；cache.put（T，hi）；cache.put（newLong（II），hey）；Thiscodeinsertsthreeobjectsintothecache.Why？（）A.hashCode（）andequals（）methodforeachobjecttypeisdifferentsoadifferentkeyisusedB.eachobjectvaluestringisdifferentsoadifferentvalueisinsertedoneachputC.equals（）andcompare（）methodisdifferentforeachputD.POFneedtobeimplementedforthistoworkproperly
1、哲学家就餐问题的解决方案如下： semephore *chopstick[5]; semaphore *seat; 哲学家 i: …… P（seat); P（chopStick[i]); P（chopStick[（i + 1) % 5]); 吃饭 V（chopStick[i]); V（chopStick[（i + 1) % 5]); V（seat); 其中，信号量seat的初值为（）。A．0B．1C．4D．5
下面的代码用于输出字符数组ch中每个字符出现的次数，应该填入的代码是（）public static void main（String[] args) { char[] ch = { 'a', 'c', 'a', 'b', 'c', 'b' }; HashMap map = new HashMap（); for （int i = 0; i ＜ ch.length; i++) { ＜ 填入代码 ＞ } System.out.println（map); }A．if (map.contains(ch[i])) { map.put(ch[i], map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }B．if (map.contains(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }C．if (map.containsKey(ch[i])) { map.put(ch[i], (int) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }D．if (map.containsKey(ch[i])) { map.put(ch[i], (Integer) map.get(ch[i]) + 1); } else { map.put(ch[i], 1); }
哲学家就餐问题的解决方案如下： semephore *chopstick[5]; semaphore *seat; 哲学家 i: …… P（seat); P（chopStick[i]); P（chopStick[（i + 1) % 5]); 吃饭 V（chopStick[i]); V（chopStick[（i + 1) % 5]); V（seat); 其中，信号量seat的初值为（）。A．0B．1C．4D．5