单选题Lifeboats are numbered（）.Odd numbers to starboard and even numbers to port.A from forward to aftB from aft to forwardC at master's optionD as per owner's instruction

Your vessel has lifeboats on both sides. Lifeboat No. 2 is located ________.

A.forward of lifeboat No. 4 on the starboard side

B.forward of lifeboat No. 4 on the port side

C.aft of lifeboat No. 1 on the starboard side

D.All of the above

函数ReadDat()实现从文件IN.dat中读取1000个十进制整数到数组xx中；请编写函数 Compute()分别计算出数组xx中奇数的个数odd、偶数的个数even，以及所有数的平均值aver和方差totfc的值。最后调用函数WriteDat()，把结果输出到OUT. dat文件中。

计算方差的公式如下：

原始数据文件存放的格式是：每行存放10个数(每个数均大于0且小于等于2000)，并用逗号隔开。

注意：部分源程序已经给出。请勿改动主函数main()、读函数ReadDat()和输出数据函数 WriteDat()的内容。

include ＜stdio. h＞

include ＜stdlib. h＞

include ＜string. h＞

define MAX 1000

int xx[MAX], odd = 0, even = 0;

double aver = 0.0, totfc = 0.0;

void WriteDat(void);

int ReadDat(void)

{ FILE *fp;

int i, j;

if ((fp = fopen("IN. dat", "r")) == NULL)

return 1;

for (i=0; i＜100; i++)

{ for (j=0; j＜10; j++)

fscanf(fp, "%d,", &xx[i*10+j]);

fscanf(fp, "\n");

if (feof(fp))

break;

}

fclose(fp);

return 0;

}

void Compute(void)

{

}

void main ()

{ int i;

for (i=0; i＜MAX; i++)

xx[i] = 0;

if (ReadDat())

{

printf (" 数据文件 IN. dar 不能打开! \007\n");

return;

}

Compute ( );

printf("odd=%d\neven=%d\naver=%f\ntotfc=%f\n", odd, even, aver, totfc);

WriteDat ();

}

void WriteDat(void)

{ FILE *fp;

fp = fopen("OUT.dat", "w");

fprintf(fp, "%d\n%d\n%f\n%f\n", odd, even, aver, totfc);

fclose (fp);

}

对输入一个整数进行判断，若是偶数，输出even，否则输出odd，在子函数fun2功能是判定整数是否为偶数，若是偶数，返回1，否则返回0int fun2(int x){if(x%2==0)；return 0;}main(){int n;scanf("%d",&n);ifprintf("even\n");else printf(odd\n");}

Lifeboats are numbered ______.Odd numbers to starboard and even numbers to port．

A.from forward to aft

B.from aft to forward

C.at master's option

D.as per owner's instruction

A.Compilationfails.

B.“odd”willalwaysbeoutput.

C.“even”willalwaysbeoutput.

D.“odd”willbeoutputforoddvaluesofx，and“even”forevenvalues.

E.“even”willbeoutputforaddvaluesofx，and“odd”forevenvalues.

请编制函数ReadDat()实现从文件IN93.DAT中读取1000个十进制整数到数组xx中。请编制函数 Compute()分别计算出xx中奇数的个数odd、偶数的个数even、平均值aver及方差totfc的值，最后调用函数WriteDat()把结果输出到OUT93.DAT文件中。

计算方差的公式如下：

原始数据的存放格式是：每行存放10个数，并用逗号隔开(每个数均大于0且小于等于2000)。

注意：部分源程序已给出。

请勿改动主函数main()和写函数WriteDat()的内容。

试题程序：

include＜stdio.h＞

include＜stdlib.h＞

include＜string.h＞

define MAX 1000

int xx[MAX],odd=0,even=0;

double aver=0.0,totfc=0.0;

void WriteDat(void);

int ReadDat(void)

{

FILE *fp;

if((fp=fopen("IN93.DAT","r"))==NULL

return 1;

fclose(fp);

return O;

}

void Compute(void)

{

}

void main( )

{

int i;

for(i=O;i＜MAX;i++)

xx[i] =0;

if (ReadDat ( ) )

{

print f ("数据文件IN93.DAT不能打开! \007\n");

return;

}

Compute();

printf( "ODD=%d\nEVEN=%d\nAVER=%lf\nTOTFC=%lf\n",

odd, even,aver,totfc);

WriteDat();

}

void WriteDat(void)

{

FILE *fp;

int i;

fp=fopen("OUT93.DAT","w");

fprintf(fp,"%dln%dln%lf\n%lfkn",odd,even,aver,totfc);

fclose(fp);

}

Preferred channel buoys indicate the preferred channel to transit by ______.

A.odd or even numbers

B.the color of their top band

C.the location of the buoy in the channel junction

D.the buoy’s light rhythms