實驗7 ?一維數(shù)組、二維數(shù)組及字符串?dāng)?shù)組

任務(wù)1 ?數(shù)組的基本定義與應(yīng)用

1．4 ??3 ??3 ??2

2．6,4,3,2,

3．a=a+arr[i][j]

???j=2

???j--

i+j==2

輸出為：a=13,b=14

4．i=1

???x[i-1]

5．m<1000或m<=999

???m/10%10或m/10-x*10

???a[i]=m

???m<i或m<=i-1

???輸出結(jié)果為：153 ??370 ??371 ??407

6. a

???a

???sum/n

???x[i]<ave

任務(wù)2 ?字符數(shù)組

1．‘\0’

???str1[i]-str2[i]

2．c[k]=a[i++];

???c[k]=b[j++];

???a[i]== ‘\0’或b[j]!= ‘\0’

3．xWHOwho

4．c1!=?‘ ’&&c2==‘ ’

實驗8 ?數(shù)組與函數(shù)

任務(wù)1 ?數(shù)組與函數(shù)的綜合應(yīng)用

蜄z ?n%2+‘0’

bin[i]=?‘\0’

2．float a[10],x;

???i<10或i<=9

???i<10或i<=9

??j<9或j<=8或j<10-i或j<=9-i

???a[j]>a[j+1]

???a[j]=a[j+1]

???i<10或i<=9

???i%5==0

3．i=strlen(a)-1;i>=j;i—

???a[i+1]=a[i]

4．m[i]=a%10

???t=t*10

???k==n*n

5．0

???||

???1

6．k-1

???N-1

???temp

7．程序代碼如下：

#include<stdio.h>

void fun(char *a)

{ ?int i=0;

???char *p=a;

???while(*p&&*p=='*')

???{a[i]=*p;

????i++;

p++;

???}

???while(*p)

???{

???if(*p!='*')

???{a[i]=*p;

?????????i++;

???}

???p++;

???}

???a[i]='\0';

}

main()

{ ?char s[81];

???printf("Enter a string:\n");

???gets(s);

???fun(s);

???printf("The string after deleted:\n");

???puts(s);

}