请大家帮忙补充一下这四个程序
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祝大家开心!
1、以下程序求a数组中所有素数之和，函数isprime用来判断自变量是否为素数，素数x是能被1和它本身整除大于1的自然数。请补充完整程序
#include "____①________"
main()
{int i, a[____②___],*p=a, sum=____③______;
Printf("Enter 10 num:\n");
for(i=0; i<10; i++) scanf("%d",___④_____);
if(isprime(*(p+____⑤_____))==1)
{ printf("&d",*(a+i));
sum+=*(a+i);
}
printf("\n The sum=%d\n",__⑥___)；
}
isprime(int x)
{ int___⑦_____；
for(i=2;___⑧_____;i++)
if(x%i==0) return(0);
if(x>1) return(1);
else return(0);
}
⑵以下程序为结构数组接收10个人的姓名、年龄和性别。然后输出，请补充完整程序
Struc man
{_①___name[20];
unsigned age;
char sex[7];
};
main()
{___②__person[10];
data_in(person);
data_out(__③____);
}
data_in(p)
Struct map *p;
{ struct man *q=__④______;
for(;p<q;p++)
{Printf("age;sex;name");
scanf("%u%s",p→age,p→sex);
___⑤____;
}
}
data_out(p)
__⑥_____*p;
{Struct man *q=___⑦___;
for(;p<q;p++)
Printf("%s;%u;__⑧____\n",p→age,p→sex);
}
⑶以下程序对10个字符串由小到大的顺序排列。main函数对字符串赋初值并输出排序结果。（其中strcmp（srt1，srt2）为字符串比较函数；若srt1<srt2 则函数值为一负数;若srt1=srt2 则函数值为0；若srt1>srt2 则函数值为一正数）
#include "stdio.h"
main()
{void sort(char *s[],int n);
char *str[10]={"java","prolog","pl","lisp","logo","VB","fortran","cobol","pascal","c++"};
int i;
sort(__①____);
Printf("new seguence of strigs is:\n");
for(___②___;___③___;___④________)
Printf("%s\n",str[i]);
}
Void sort(char *s[],int n)
{char *temp;
int i,j;
for(i=0;i<n;i++)
for(i=0;j<n;j++)
if(strcmp(s[i],s[j])<0)
{_⑤___=s[i];
__⑥__=__⑦__;
s[j]=__⑧__;}
}
⑷以下f函数功能是：在带有头结点单向链表中，查找结点数据域的最大值作为函数返回值
Struct book
Struct book
{int a;
__①___*next;
}
int f(struct book *pl)
{___②___*p2;
p2=p1→next
m=p2→a;
for(p2=p2→next;p2!='\0';p2=__③___)
if(___④____)m=p2→a;
return__⑤___;
}
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(1) 1、 stdio.h 2、10 3、0 4、&a[i] 5、i 6、sum 7、i 8、i<x
----------------解决方案--------------------------------------------------------
#include "stdio.h"
main()
{void sort(char *s[],int n);
char *str[10]={"java","prolog","pl","lisp","logo","VB","fortran","cobol","pascal","c++"};
int i;
sort(str,10);
printf("new seguence of strigs is:\n");
for(i=0;i<10;i++)
printf("%s\n",str[i]);
}
void sort(char *s[],int n)
{char *temp;
int i,j;
for(i=0;i<n;i++)
for(j=0;j<n;j++)
if(strcmp(s[i],s[j])<0)
{temp=s[i];
s[i]=s[j];
s[j]=temp;
}
}
----------------解决方案--------------------------------------------------------
struct book
{int a;
struct book *next;
}
int f(struct book *pl)
{struct book *p2;
p2=p1→next
m=p2→a;
for(p2=p2→next;p2!='\0';p2=p2->next)
if(p2->a>m)m=p2→a;
return m;
}
----------------解决方案--------------------------------------------------------
对于程序1：
#include "stdio.h"
#include "math.h"
main()
{int i, a[10],*p=a, sum=0;
Printf("Enter 10 num:\n");
for(i=0; i<10; i++) scanf("%d",a+i);
if(isprime(*(p+i))==1)
{ printf("&d",*(a+i));
sum+=*(a+i);
}
printf("\n The sum=%d\n",sum)；
}
isprime(int x)
{ int i；
for(i=2;i<=(int) sqrt(x);i++)
if(x%i==0) return(0);
if(x>1) return(1);
else return(0);
}
----------------解决方案--------------------------------------------------------
谢谢大家的帮助啊!
----------------解决方案--------------------------------------------------------