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<div align="center"><font color="#FFFF00" size="5">经典c程序100例==61--70 </font></div>
<p>【程序61】<br>
题目：打印出杨辉三角形（要求打印出10行如下图） <br>
1.程序分析：<br>
1<br>
1 1<br>
1 2 1<br>
1 3 3 1<br>
1 4 6 4 1<br>
1 5 10 10 5 1 <br>
2.程序源代码：<br>
main()<br>
{int i,j;<br>
int a[10][10];<br>
printf("\n");<br>
for(i=0;i<10;i++)<br>
{a[i][0]=1;<br>
a[i][i]=1;}<br>
for(i=2;i<10;i++)<br>
for(j=1;j<i;j++)<br>
a[i][j]=a[i-1][j-1]+a[i-1][j];<br>
for(i=0;i<10;i++)<br>
{for(j=0;j<=i;j++)<br>
printf("%5d",a[i][j]);<br>
printf("\n");<br>
}<br>
}<br>
==============================================================<br>
【程序62】<br>
题目：学习putpixel画点。<br>
1.程序分析： <br>
2.程序源代码：<br>
#include "stdio.h"<br>
#include "graphics.h"<br>
main()<br>
{<br>
int i,j,driver=VGA,mode=VGAHI;<br>
initgraph(&driver,&mode,"");<br>
setbkcolor(YELLOW);<br>
for(i=50;i<=230;i+=20)<br>
for(j=50;j<=230;j++)<br>
putpixel(i,j,1);<br>
for(j=50;j<=230;j+=20)<br>
for(i=50;i<=230;i++)<br>
putpixel(i,j,1);<br>
}<br>
==============================================================<br>
【程序63】<br>
题目：画椭圆ellipse <br>
1.程序分析：<br>
2.程序源代码：<br>
#include "stdio.h"<br>
#include "graphics.h"<br>
#include "conio.h"<br>
main()<br>
{<br>
int x=360,y=160,driver=VGA,mode=VGAHI;<br>
int num=20,i;<br>
int top,bottom;<br>
initgraph(&driver,&mode,"");<br>
top=y-30;<br>
bottom=y-30;<br>
for(i=0;i<num;i++)<br>
{<br>
ellipse(250,250,0,360,top,bottom);<br>
top-=5;<br>
bottom+=5;<br>
}<br>
getch();<br>
}<br>
==============================================================<br>
【程序64】<br>
题目：利用ellipse and rectangle 画图。<br>
1.程序分析：<br>
2.程序源代码： <br>
#include "stdio.h"<br>
#include "graphics.h"<br>
#include "conio.h"<br>
main()<br>
{<br>
int driver=VGA,mode=VGAHI;<br>
int i,num=15,top=50;<br>
int left=20,right=50;<br>
initgraph(&driver,&mode,"");<br>
for(i=0;i<num;i++)<br>
{<br>
ellipse(250,250,0,360,right,left);<br>
ellipse(250,250,0,360,20,top);<br>
rectangle(20-2*i,20-2*i,10*(i+2),10*(i+2));<br>
right+=5;<br>
left+=5;<br>
top+=10;<br>
}<br>
getch();<br>
}<br>
==============================================================<br>
【程序65】<br>
题目：一个最优美的图案。 <br>
1.程序分析：<br>
2.程序源代码：<br>
#include "graphics.h"<br>
#include "math.h"<br>
#include "dos.h"<br>
#include "conio.h"<br>
#include "stdlib.h"<br>
#include "stdio.h"<br>
#include "stdarg.h"<br>
#define MAXPTS 15<br>
#define PI 3.1415926<br>
struct PTS {<br>
int x,y;<br>
};<br>
double AspectRatio=0.85;<br>
void LineToDemo(void)<br>
{<br>
struct viewporttype vp;<br>
struct PTS points[MAXPTS];<br>
int i, j, h, w, xcenter, ycenter;<br>
int radius, angle, step;<br>
double rads;<br>
printf(" MoveTo / LineTo Demonstration" );<br>
getviewsettings( &vp );<br>
h = vp.bottom - vp.top;<br>
w = vp.right - vp.left;<br>
xcenter = w / 2; /* Determine the center of circle */<br>
ycenter = h / 2;<br>
radius = (h - 30) / (AspectRatio * 2);<br>
step = 360 / MAXPTS; /* Determine # of increments */<br>
angle = 0; /* Begin at zero degrees */<br>
for( i=0 ; i<MAXPTS ; ++i ){ /* Determine circle intercepts */<br>
rads = (double)angle * PI / 180.0; /* Convert angle to radians */<br>
points[i].x = xcenter + (int)( cos(rads) * radius );<br>
points[i].y = ycenter - (int)( sin(rads) * radius * AspectRatio );<br>
angle += step; /* Move to next increment */<br>
}<br>
circle( xcenter, ycenter, radius ); /* Draw bounding circle */<br>
for( i=0 ; i<MAXPTS ; ++i ){ /* Draw the cords to the circle */<br>
for( j=i ; j<MAXPTS ; ++j ){ /* For each remaining intersect */<br>
moveto(points[i].x, points[i].y); /* Move to beginning of cord */<br>
lineto(points[j].x, points[j].y); /* Draw the cord */<br>
} } }<br>
main()<br>
{int driver,mode;<br>
driver=CGA;mode=CGAC0;<br>
initgraph(&driver,&mode,"");<br>
setcolor(3);<br>
setbkcolor(GREEN);<br>
LineToDemo();}<br>
==============================================================<br>
【程序66】<br>
题目：输入3个数a,b,c，按大小顺序输出。 <br>
1.程序分析：利用指针方法。<br>
2.程序源代码：<br>
/*pointer*/<br>
main()<br>
{<br>
int n1,n2,n3;<br>
int *pointer1,*pointer2,*pointer3;<br>
printf("please input 3 number:n1,n2,n3:");<br>
scanf("%d,%d,%d",&n1,&n2,&n3);<br>
pointer1=&n1;<br>
pointer2=&n2;<br>
pointer3=&n3;<br>
if(n1>n2) swap(pointer1,pointer2);<br>
if(n1>n3) swap(pointer1,pointer3);<br>
if(n2>n3) swap(pointer2,pointer3);<br>
printf("the sorted numbers are:%d,%d,%d\n",n1,n2,n3);<br>
}<br>
swap(p1,p2)<br>
int *p1,*p2;<br>
{int p;<br>
p=*p1;*p1=*p2;*p2=p;<br>
}<br>
==============================================================<br>
【程序67】<br>
题目：输入数组，最大的与第一个元素交换，最小的与最后一个元素交换，输出数组。<br>
1.程序分析：谭浩强的书中答案有问题。 <br>
2.程序源代码：<br>
main()<br>
{<br>
int number[10];<br>
input(number);<br>
max_min(number);<br>
output(number);<br>
}<br>
input(number)<br>
int number[10];<br>
{int i;<br>
for(i=0;i<9;i++)<br>
scanf("%d,",&number[i]);<br>
scanf("%d",&number[9]);<br>
}<br>
max_min(array)<br>
int array[10];<br>
{int *max,*min,k,l;<br>
int *p,*arr_end;<br>
arr_end=array+10;<br>
max=min=array;<br>
for(p=array+1;p<arr_end;p++)<br>
if(*p>*max) max=p;<br>
else if(*p<*min) min=p;<br>
k=*max;<br>
l=*min;<br>
*p=array[0];array[0]=l;l=*p;<br>
*p=array[9];array[9]=k;k=*p;<br>
return;<br>
}<br>
output(array)<br>
int array[10];<br>
{ int *p;<br>
for(p=array;p<array+9;p++)<br>
printf("%d,",*p);<br>
printf("%d\n",array[9]);<br>
}<br>
==============================================================<br>
【程序68】<br>
题目：有n个整数，使其前面各数顺序向后移m个位置，最后m个数变成最前面的m个数<br>
1.程序分析：<br>
2.程序源代码：<br>
main()<br>
{<br>
int number[20],n,m,i;<br>
printf("the total numbers is:");<br>
scanf("%d",&n);<br>
printf("back m:");<br>
scanf("%d",&m);<br>
for(i=0;i<n-1;i++)<br>
scanf("%d,",&number[i]);<br>
scanf("%d",&number[n-1]);<br>
move(number,n,m);<br>
for(i=0;i<n-1;i++)<br>
printf("%d,",number[i]);<br>
printf("%d",number[n-1]);<br>
}<br>
move(array,n,m)<br>
int n,m,array[20];<br>
{<br>
int *p,array_end;<br>
array_end=*(array+n-1);<br>
for(p=array+n-1;p>array;p--)<br>
*p=*(p-1);<br>
*array=array_end;<br>
m--;<br>
if(m>0) move(array,n,m);<br>
}<br>
==============================================================<br>
【程序69】<br>
题目：有n个人围成一圈，顺序排号。从第一个人开始报数（从1到3报数），凡报到3的人退出<br>
圈子，问最后留下的是原来第几号的那位。<br>
1. 程序分析：<br>
2.程序源代码：<br>
#define nmax 50<br>
main()<br>
{<br>
int i,k,m,n,num[nmax],*p;<br>
printf("please input the total of numbers:");<br>
scanf("%d",&n);<br>
p=num;<br>
for(i=0;i<n;i++)<br>
*(p+i)=i+1;<br>
i=0;<br>
k=0;<br>
m=0;<br>
while(m<n-1)<br>
{<br>
if(*(p+i)!=0) k++;<br>
if(k==3)<br>
{ *(p+i)=0;<br>
k=0;<br>
m++;<br>
}<br>
i++;<br>
if(i==n) i=0;<br>
}<br>
while(*p==0) p++;<br>
printf("%d is left\n",*p);<br>
}<br>
==============================================================<br>
【程序70】<br>
题目：写一个函数，求一个字符串的长度，在main函数中输入字符串，并输出其长度。 <br>
1.程序分析：<br>
2.程序源代码：<br>
main()<br>
{<br>
int len;<br>
char *str[20];<br>
printf("please input a string:\n");<br>
scanf("%s",str);<br>
len=length(str);<br>
printf("the string has %d characters.",len);<br>
}<br>
length(p)<br>
char *p;<br>
{<br>
int n;<br>
n=0;<br>
while(*p!='\0')<br>
{<br>
n++;<br>
p++;<br>
}<br>
return n;<br>
}<br>
</p>
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