1994年程序员下午考试
发表于:2007-05-26来源:作者:点击数:
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【程 序】 #define MAXSCORE 20 #define QUESTION 10 #define ORDERS 5 main() { int p[QUESTION]={0,0,0,0,0,0,0,0,0,0}, n[QUESTION]={0,0,0,0,0,0,0,0,0,0}, s[QUESTION]={0,0,0,0,0,0,0,0,0,0}; int f[ORDERS]={0,0,0,0,0}; int i,score,c,number,pn=0;
【程 序】
#define MAXSCORE 20
#define QUESTION 10
#define ORDERS 5
main()
{ int p[QUESTION]={0,0,0,0,0,0,0,0,0,0},
n[QUESTION]={0,0,0,0,0,0,0,0,0,0},
s[QUESTION]={0,0,0,0,0,0,0,0,0,0};
int f[ORDERS]={0,0,0,0,0};
int i,score,c,number,pn=0;
char fig,ch[120];
char *title[]={" 90 -- 100 A",
" 80 -- 89 B",
" 70 -- 79 C",
" 60 -- 69 D",
" 0 -- 59 E"}
while(1)
{
printf("Enter number && score1 -- score10 \n");
if (scanf("%d",&number) ==0)
{
gets(ch);
printf("Error! Input again!\n");
continue;
}
for (c=0,i=1;i<QUESTION && c== i; i++)
if (scanf("%d",&p
))
if (p <= MAXSCORE)
_________________________ ;
if ( ______________________ )
{
gets(ch);
printf("Error! Input again!\n");
continue;
}
for (c=0,score=0,i=0;i<QUESTION;i++)
if ( _______________ )
{
c++; score +=p; n++; s +=p;
}
fig = (score ==100) ? 'A': (score < 60) ? _____________________;
f[ _______ ]++; pn++;
printf("Number = %d Score = %d Mark = %c\n",number,score,fig);
}
printf("STUDENTS = %d\n",pn);
for (i=0;i<ORDERS;i++) printf("%s%7d\n",title,f);
printf("\n Question Students Average\n");
for (i=0;i<QUESTION;i++)
if (n) printf("%6d%10d%10.2f\n",i+1,n, _______________ );
else pritnf ("6d%10d%10s\n",i+1,n," --");
}
本程序实现安照每页宽80列平均分左右两栏的格式
印出正文文件内容.
程序引入数组buff[] [] [] 和ln [] [], 将从文件
读出的字符按行存储于buff[0],行号存于ln[0](对应左栏
), 或buff[1],ln[1](对应右栏).约定,文件内容先填左栏
填满后,再填右栏.或左右两栏填满,或文件内容填完,输出
一页的内容.
欲输出的正文文件(小于1000行)的文件名作为主函数
的参数.主函数以文件名为参数调用函数dprint()输出一个
文件.函数dprint()读取文件内容,控制栏中的一行内容的
填写,当一行填满时,调用函数nextline().函数nextline()
控制栏中行的变化,左右栏的变化.待左右栏都填满时,调用
函数printout()完成整页输出.函数printout()完成页面排
版,取ln[0] buff[0]和ln[1] buff[1],将对应行号及内容
填入line[],逐行输出.
【程 序】
#include <stdio.h>
#define LL 80
#define COL 2
#define CSIZE LL/COL-9
#define PL 50
#define MARGIN 3
char buff[COL][PL][CSIZE];
int ln[COL][PL];
int col,row,p;
dprint(char *fname)
{ FILE *fp;
int lin,c;
if ((fp=fopen(fname,"r"))==NULL) return;
lin =0; p=0; col=0; c=getc(fp);
while (c!=EOF)
{ ln[col][row]=++lin;
while (c != '\n' && c != EOF)
{ if (p>= CSIZE)
{
_________________ ; ln[col][row] = 0;
}
_________________ = c ; c = getc(fp);
}
____________________ ;
if (c != EOF) c = getc(fp);
}
while( col != 0 || row != 0 )
{ ln[col][row] = 0;
nextline();
}
fclose(fp);
}
nextline()
{ while(p < CSIZE)
buff[col][row][p++] = ' ';
if ( _____________ )
{ if ( ++col >= COL )
{ printout();
_______________;
}
row = 0;
}
p = 0;
}
printout()
{ int k, i, lpos, col, d;
char line[LL];
for(k=0;k<MARGIN;k++) putchar('\n');
for(k=0;k<PL;k++)
{ for(i=0;i<LL-1;i++)
for(lpos=0,col=0;col<COL;lpos += CSIZE+9,
col++)
{ d = _____________;
p = lpos + 4;
while (d>0)
{ line[p--] = _______________;
d /= 10;
}
for(p=lpos+7,i=0;i<CSIZE;i++)
line[p++] = buff[col][k];
}
puts(line);
}
for(k=0;k<MARGIN;k++) putchar('\n');
}
main(int argc, char **argv)
{ int f;
for(f=1;f<argc;f++)
dprint(argc[f]);
}
本程序给出两个函数.函数create()根据已知整数数组构造一个线性链表.函
数sort()采用选择排序方法对已知链表进行排序.为排序方便,函数sort()于排
序前在链表首表元之前生成一个辅助表元.排序完成后,将该辅助表元筛去.
【程 序】
#include <stdio.h>
#include <stdlib.h>
struct node{
int value;
struct node *next;
};
struct node *create(int a[], int n)
{ struct node *h, *q;
for(h=NULL;n;n--)
{ q = (struct node *)malloc(sizeof(struct node));
q->value = ____________;
______________;
______________;
}
return h;
}
void sort(struct node **h)
{ struct node *p,*q,*r,*s,*hl;
hl = p = (struct node*)malloc(sizeof(struct node));
p->next = *h;
while(p->next != NULL)
{ q = p->next;
r = p;
while(p->next != NULL)
{ if (q->next->value < __________ )
r = q;
q = q->next;
}
if( r != p )
{ s = ____________;
_____________ = s->next;
s->next = ___________;
___________ = s;
}
p = p->next;
}
*h = hl->next;
free(hl);
}
int text_data[] = {5,9,3,4,5,7,8};
main()
{ struct node *h, *p;
h = create(test_data,
sizeof test_data/size of test_data);
for(p=h;p;p=p->next) printf("%5d",p->value);
printf("\n");
sort(&h);
for(p=h;p;p=p->next) printf("%5d",p->value);
printf("\n");
}
原文转自:http://www.ltesting.net