问答题A jar contains 30 coins, of which 20 are nickels and 10 are dimes. If 7 coins are removed, how many nickels are left?　　(1) More nickels are removed than dimes.　　(2) The remaining coins in the jar are worth $1.55.

18 of the EU members have replaced their national currencies by Euro notes and coins ()2002.

A. for

B. at

C. since

A.If line 10 is removed, the compilation succeeds.

B.If line 11 is removed, the compilation succeeds.

C.If line 12 is removed, the compilation succeeds.

D.If line 13 is removed, the compilation succeeds.

E.More than one line must be removed for compilation to succeed.

【T1】

A.TO OBTAIN

B.MADE OF

C.IS WORTH A.IT IS CHEAPER【T1】______SILVER

B.THE COINS WERE ACTUALLY【T2】______SILVER

C.THE SILVER IN COINS【T3】______MUCH MORE LESS THAN 40 YEARS AGO IN THE UNITED STATES, IT WAS COMMON TO CHANGE A ONE-DOLLAR BILL FOR A DOLLAR"S WORTH OF SILVER.THAT IS BECAUSE【T4】______.BUT THOSE DAYS ARE GON

E.THERE IS NO SILVER IN TODAY"S COINS.WHEN THE PRICE OF THE PRECIOUS METAL RISES ABOVE ITS FACE VALUE AS MONEY, THE METAL WILL BECOME MORE VALUABLE IN OTHER USES.SILVER COINS ARE NO LONGER IN CIRCULATION BECAUSE【T5】______THAN THEIR FACE VALU

E.A SILVER FIRM COULD FIND THAT【T6】______BY MELTING DOWN COINS THAN BY BUYING IT ON THE COMMODITY MARKETS.COINS TODAY ARE MADE OF AN ALLOY OF CHEAPER METALS.

【T3】

A.TO OBTAIN

B.MADE OF

C.IS WORTH A.IT IS CHEAPER【T1】______SILVER

B.THE COINS WERE ACTUALLY【T2】______SILVER

C.THE SILVER IN COINS【T3】______MUCH MORE LESS THAN 40 YEARS AGO IN THE UNITED STATES, IT WAS COMMON TO CHANGE A ONE-DOLLAR BILL FOR A DOLLAR"S WORTH OF SILVER.THAT IS BECAUSE【T4】______.BUT THOSE DAYS ARE GON

E.THERE IS NO SILVER IN TODAY"S COINS.WHEN THE PRICE OF THE PRECIOUS METAL RISES ABOVE ITS FACE VALUE AS MONEY, THE METAL WILL BECOME MORE VALUABLE IN OTHER USES.SILVER COINS ARE NO LONGER IN CIRCULATION BECAUSE【T5】______THAN THEIR FACE VALU

E.A SILVER FIRM COULD FIND THAT【T6】______BY MELTING DOWN COINS THAN BY BUYING IT ON THE COMMODITY MARKETS.COINS TODAY ARE MADE OF AN ALLOY OF CHEAPER METALS.

Click the Exhibit button. Given the fully-qualified class names: com.foo.bar.Dog com.foo.bar.blatz.Book com.bar.Car com.bar.blatz.Sun Which graph represents the correct directory structure for a JAR file from which those classes can be used by the compiler and JVM? （）

A.Jar A

B.Jar B

C.Jar C

D.Jar D

E.Jar E

使用VC6打开考生文件夹下的工程test36_1，此工程包含一个源程序文件test36_1.cpp，但该程序运行有问题，请改正函数中的错误，使该程序的输出结果为：

2

源程序文件test36_1.cpp清单如下：

include＜iostream.h＞

class amount;

class coins

{

/***************** found *****************/

enum units {penny, nickel, dime, quarter, half_dollar}

/***************** found *****************/

class amount;

};

class amount

{

/***************** found *****************/

coins:units money;

public:

void setm();

int germ();

};

void amount::setm()

{

money = coins::dime;

}

int amount::getm()

{

return money;

}

int main ( )

{

amount ob;

ob.setm();

cout ＜＜ ob.getm()＜＜end1;

return 0;

}

【T2】

A.TO OBTAIN

B.MADE OF

C.IS WORTH A.IT IS CHEAPER【T1】______SILVER

B.THE COINS WERE ACTUALLY【T2】______SILVER

C.THE SILVER IN COINS【T3】______MUCH MORE LESS THAN 40 YEARS AGO IN THE UNITED STATES, IT WAS COMMON TO CHANGE A ONE-DOLLAR BILL FOR A DOLLAR"S WORTH OF SILVER.THAT IS BECAUSE【T4】______.BUT THOSE DAYS ARE GON

E.THERE IS NO SILVER IN TODAY"S COINS.WHEN THE PRICE OF THE PRECIOUS METAL RISES ABOVE ITS FACE VALUE AS MONEY, THE METAL WILL BECOME MORE VALUABLE IN OTHER USES.SILVER COINS ARE NO LONGER IN CIRCULATION BECAUSE【T5】______THAN THEIR FACE VALU

E.A SILVER FIRM COULD FIND THAT【T6】______BY MELTING DOWN COINS THAN BY BUYING IT ON THE COMMODITY MARKETS.COINS TODAY ARE MADE OF AN ALLOY OF CHEAPER METALS.

阅读下列说明和C代码，回答问题 1 至问题 3，将解答写在答题纸的对应栏内。 【说明】 假币问题：有n枚硬币，其中有一枚是假币，己知假币的重量较轻。现只有一个天平，要求用尽量少的比较次数找出这枚假币。 【分析问题】 将n枚硬币分成相等的两部分: (1)当n为偶数时，将前后两部分，即 1...n/2和n/2+1...0，放在天平的两端，较轻的一端里有假币，继续在较轻的这部分硬币中用同样的方法找出假币: (2)当n为奇数时，将前后两部分，即1..(n -1)/2和(n+1)/2+1...0，放在天平的两端，较轻的一端里有假币，继续在较轻的这部分硬币中用同样的方法找出假币；若两端重量相等，则中间的硬币，即第 (n+1)/2枚硬币是假币。 【C代码】 下面是算法的C语言实现，其中: coins[]： 硬币数组 first，last：当前考虑的硬币数组中的第一个和最后一个下标 include <stdio.h> int getCounterfeitCoin(int coins[]， int first，int last) { int firstSum = 0，lastSum = 0; int ì; If(first==last-1){ /*只剩两枚硬币*/ if(coins[first] < coins[last]) return first; return last; } if((last - first + 1) % 2 ==0){ /*偶数枚硬币*/ for(i = first;i <( 1 );i++){ firstSum+= coins[i]; } for(i=first + (last-first) / 2 + 1;i < last +1;i++){ lastSum += coins[i]; } if( 2 ){ Return getCounterfeitCoin(coins,first,first+(last-first)/2;) }else{ Return getCounterfeitCoin(coins,first+(last-first)/2+1,last;) } } else{ /*奇数枚硬币*/ For(i=first;i<first+(last-first)/2;i++){ firstSum+=coins[i]; } For(i=first+(last-first)/2+1;i<last+1;i++){ lastSum+=coins[i]; } If(firstSum<lastSum){ return getCounterfeitCoin(coins,first,first+(last-first)/2-1); }else if(firstSum>lastSum){ return getCounterfeitCoin(coins,first+(last-first)/2-1,last); }else{ Return( 3 ) } } }

【问题一】 根据题干说明，填充C代码中的空（1）-（3） 【问题二】 根据题干说明和C代码，算法采用了（ ）设计策略。 函数getCounterfeitCoin的时间复杂度为（ ）（用O表示）。 【问题三】 若输入的硬币数为30，则最少的比较次数为（ ），最多的比较次数为（ ）。