I have two leaves.()
I have two leaves.()
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3． —Come and join us，Jimmy !—I-m sorry，but I-m really busy now．lf I _______ time，l would certainly go．A． will haveB． have hadC． hadD． have
And they can ______their passports______，making leaving or “escaping” virtually impossible.A. have…removeB. had…removeC. have…removingD. have…removed
（ 10 ）执行下列程序，显示的结果是 【 10 】 。one= "WORK"two = ""a = LEN ( one )i = aDO WHILE i=1two = two + SUBSTR ( one ,i , 1 ）i=i - 1ENDDO? two
What a waste! Why do you () the water () all the time ? A、have…flowingB、have…flowC、have…to flowD、have…flowed
执行下列程序，显示的结果是【 】。ne="WORK"two=""a=LEN(one)i=aDO WHILE i＞=ltwo=two+SUBSTR(one,i, 1)i=i-1ENDDO?two
编译和执行以下代码，输出结果是( )。 int i=1; switch (i) { case 0: System.out.print("zero,"); break; case 1: System.out.print("one,"); case 2: System.out.print("two,"); default: System.out.println("default"); }A．one，B．one,two,C．one,two,defaultD．default
阅读以下说明及C++程序代码，将应填入(n)处的语句写在对应栏内。【说明】本程序的功能是实现任意两个大整数的乘法运算，例如：输入整数1：8934793850094505800243958034985058输入整数2：234584950989689084095803583095820923二者之积：209596817742739508050978890737675662366433464256830959194834854876 8534【C++代码】include＜iostream.h＞const int MAXINPUTBIT=100;const int MAXRESULTBIT=500;class LargeNumber{int i,j;int temp;int one[MAXINPUTBIT+1];int onebit; //one的位数int two[MAXINPUTBIT+1];int twobit; //two的位数int result[MAXRESULTBIT+1];public:LargeNumber();~LargeNumber();int inputone(); //出错返叫0，否则返回1int inputtwo(); //同上void multiplication(); //乘void clearresult(); //清零void showresult(); //显示};LargeNumber∷LargeNumber(){for(i=0;i＜=MAXINPUTBIT;i++){one[i]=0;two[i]=0;}nebit=0;twobit=0;inputone();inputtwo();}LargeNumber∷～LargeNumber(){}int LargeNumber∷inputone(){char Number[MAXINPUTBIT+1];cout＜＜"Please enter one:";cin＞＞Number;i=0;j=MAXINPUTBIT;while(Number[i]!='\0')i++;nebit=i;for(i--;i＞=0;i--,j--){if(int(Number[i])＞=48int(Number[i])＜=57)(1); //由字符转换为数字elsereturn 0;}return 1;}int LargeNumber∷inputtwo(){char Number[MAXINPUTBIT+1];cout＜＜"Please enter two:";cin＞＞Number;i=0;j=MAXINPUTBIT;while(Number[i]!='\0')i++;twobit=i;for(i--;i＞=0;i--,j--){if(int(Number[i])＞=48int(Number[i])＜=57)two[j]=int(Number[i]-48); //由字符转换为数字elsereturn 0;}return 1;}void LargeNumber∷multiplication() //乘法{clearresult();int m;for(i=MAXINPUTBIT;i＞=0;i--){temp=two[i];for(j=(2),m=MAXINPUTBIT;m＞=0;m--,j--){result[j]+=temp*one[m];if(result[j]＞9){result[j-1]+=result[j]/10;(3);}}&n