計算機等級二級C語言程序設(shè)計100例六

字號:

【程序61】
    題目:打印出楊輝三角形(要求打印出10行如下圖)
    1.程序分析:
    1
    1  1
    1  2  1
    1  3  3  1
    1  4  6  4  1
    1  5  10 10 5  1
    2.程序源代碼:
    以下是引用片段:
    main()
    {int i,j;
    int a[10][10];
    printf("/n");
    for(i=0;i<10;i++)
    {a[i][0]=1;
    a[i][i]=1;}
    for(i=2;i<10;i++)
    for(j=1;j  a[i][j]=a[i-1][j-1]+a[i-1][j];
    for(i=0;i<10;i++)
    {for(j=0;j<=i;j++)
    printf("]",a[i][j]);
    printf("/n");
    }
    }
    【程序62】
    題目:學(xué)習(xí)putpixel畫點。
    1.程序分析:
    2.程序源代碼:
    以下是引用片段:
    #include "stdio.h"
    #include "graphics.h"
    main()
    {
    int i,j,driver=VGA,mode=VGAHI;
    initgraph(&driver,&mode,"");
    setbkcolor(YELLOW);
    for(i=50;i<=230;i+=20)
    for(j=50;j<=230;j++)
    putpixel(i,j,1);
    for(j=50;j<=230;j+=20)
    for(i=50;i<=230;i++)
    putpixel(i,j,1);
    }
    【程序63】
    題目:畫橢圓ellipse
    1.程序分析:
    2.程序源代碼:
    以下是引用片段:
    #include "stdio.h"
    #include "graphics.h"
    #include "conio.h"
    main()
    {
    int x=360,y=160,driver=VGA,mode=VGAHI;
    int num=20,i;
    int top,bottom;
    initgraph(&driver,&mode,"");
    top=y-30;
    bottom=y-30;
    for(i=0;i{
    ellipse(250,250,0,360,top,bottom);
    top-=5;
    bottom+=5;
    }
    getch();
    }
    【程序64】
    題目:利用ellipse and rectangle 畫圖。
    1.程序分析:
    2.程序源代碼:
    以下是引用片段:
    #include "stdio.h"
    #include "graphics.h"
    #include "conio.h"
    main()
    {
    int driver=VGA,mode=VGAHI;
    int i,num=15,top=50;
    int left=20,right=50;
    initgraph(&driver,&mode,"");
    for(i=0;i{
    ellipse(250,250,0,360,right,left);
    ellipse(250,250,0,360,20,top);
    rectangle(20-2*i,20-2*i,10*(i+2),10*(i+2));
    right+=5;
    left+=5;
    top+=10;
    }
    getch();
    }
    【程序65】
    題目:一個美的圖案。
    1.程序分析:
    2.程序源代碼:
    以下是引用片段:
    #include "graphics.h"
    #include "math.h"
    #include "dos.h"
    #include "conio.h"
    #include "stdlib.h"
    #include "stdio.h"
    #include "stdarg.h"
    #define MAXPTS 15
    #define PI 3.1415926
    struct PTS {
    int x,y;
    };
    double AspectRatio=0.85;
    void LineToDemo(void)
    {
    struct viewporttype vp;
    struct PTS points[MAXPTS];
    int i, j, h, w, xcenter, ycenter;
    int radius, angle, step;
    double rads;
    printf(" MoveTo / LineTo Demonstration" );
    getviewsettings( &vp );
    h = vp.bottom - vp.top;
    w = vp.right - vp.left;
    xcenter = w / 2; /* Determine the center of circle */
    ycenter = h / 2;
    radius = (h - 30) / (AspectRatio * 2);
    step = 360 / MAXPTS; /* Determine # of increments */
    angle = 0; /* Begin at zero degrees */
    for( i=0 ; irads = (double)angle * PI / 180.0; /* Convert angle to radians */
    points[i].x = xcenter + (int)( cos(rads) * radius );
    points[i].y = ycenter - (int)( sin(rads) * radius * AspectRatio );
    angle += step; /* Move to next increment */
    }
    circle( xcenter, ycenter, radius ); /* Draw bounding circle */
    for( i=0 ; ifor( j=i ; jmoveto(points[i].x, points[i].y); /* Move to beginning of cord */
    lineto(points[j].x, points[j].y); /* Draw the cord */
    } } }
    main()
    {int driver,mode;
    driver=CGA;mode=CGAC0;
    initgraph(&driver,&mode,"");
    setcolor(3);
    setbkcolor(GREEN);
    LineToDemo();}
    【程序66】
    題目:輸入3個數(shù)a,b,c,按大小順序輸出。
    1.程序分析:利用指針方法。
    2.程序源代碼:
    以下是引用片段:
    /*pointer*/
    main()
    {
    int n1,n2,n3;
    int *pointer1,*pointer2,*pointer3;
    printf("please input 3 number:n1,n2,n3:");
    scanf("%d,%d,%d",&n1,&n2,&n3);
    pointer1=&n1;
    pointer2=&n2;
    pointer3=&n3;
    if(n1>n2) swap(pointer1,pointer2);
    if(n1>n3) swap(pointer1,pointer3);
    if(n2>n3) swap(pointer2,pointer3);
    printf("the sorted numbers are:%d,%d,%d/n",n1,n2,n3);
    }
    swap(p1,p2)
    int *p1,*p2;
    {int p;
    p=*p1;*p1=*p2;*p2=p;
    }
    【程序67】
    題目:輸入數(shù)組,的與第一個元素交換,最小的與最后一個元素交換,輸出數(shù)組。
    1.程序分析:譚浩強的書中答案有問題。
    2.程序源代碼:
    以下是引用片段:
    main()
    {
    int number[10];
    input(number);
    max_min(number);
    output(number);
    }
    input(number)
    int number[10];
    {int i;
    for(i=0;i<9;i++)
    scanf("%d,",&number[i]);
    scanf("%d",&number[9]);
    }
    max_min(array)
    int array[10];
    {int *max,*min,k,l;
    int *p,*arr_end;
    arr_end=array+10;
    max=min=array;
    for(p=array+1;p  if(*p>*max) max=p;
    else if(*p<*min) min=p;
    k=*max;
    l=*min;
    *p=array[0];array[0]=l;l=*p;
    *p=array[9];array[9]=k;k=*p;
    return;
    }
    output(array)
    int array[10];
    { int *p;
    for(p=array;p  printf("%d,",*p);
    printf("%d/n",array[9]);
    }
    【程序68】
    題目:有n個整數(shù),使其前面各數(shù)順序向后移m個位置,最后m個數(shù)變成最前面的m個數(shù)
    1.程序分析:
    2.程序源代碼:
    以下是引用片段:
    main()
    {
    int number[20],n,m,i;
    printf("the total numbers is:");
    scanf("%d",&n);
    printf("back m:");
    scanf("%d",&m);
    for(i=0;i  scanf("%d,",&number[i]);
    scanf("%d",&number[n-1]);
    move(number,n,m);
    for(i=0;i  printf("%d,",number[i]);
    printf("%d",number[n-1]);
    }
    move(array,n,m)
    int n,m,array[20];
    {
    int *p,array_end;
    array_end=*(array+n-1);
    for(p=array+n-1;p>array;p--)
    *p=*(p-1);
    *array=array_end;
    m--;
    if(m>0) move(array,n,m);
    }
    【程序69】
    題目:有n個人圍成一圈,順序排號。從第一個人開始報數(shù)(從1到3報數(shù)),凡報到3的人退出
    圈子,問最后留下的是原來第幾號的那位。
    1. 程序分析:
    2.程序源代碼:
    以下是引用片段:
    #define nmax 50
    main()
    {
    int i,k,m,n,num[nmax],*p;
    printf("please input the total of numbers:");
    scanf("%d",&n);
    p=num;
    for(i=0;i  *(p+i)=i+1;
    i=0;
    k=0;
    m=0;
    while(m  {
    if(*(p+i)!=0) k++;
    if(k==3)
    { *(p+i)=0;
    k=0;
    m++;
    }
    i++;
    if(i==n) i=0;
    }
    while(*p==0) p++;
    printf("%d is left/n",*p);
    }
    【程序70】
    題目:寫一個函數(shù),求一個字符串的長度,在main函數(shù)中輸入字符串,并輸出其長度。
    1.程序分析:
    2.程序源代碼: 以下是引用片段:
    main()
    {
    int len;
    char *str[20];
    printf("please input a string:/n");
    scanf("%s",str);
    len=length(str);
    printf("the string has %d characters.",len);
    }
    length(p)
    char *p;
    {
    int n;
    n=0;
    while(*p!=’/0’)
    {
    n++;
    p++;
    }
    return n;
    }