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100道Python编程题及答案(四)

2023-09-12 15:40编辑: www.jxszl.com景先生毕设
           100道Python编程题及答案(四)
题目61:打印出杨辉三角形(要求打印出10行如下图)   
1.程序分析:
'''
if __name__ == '__main__':
    a = []
    for i in range(10):
        a.append([])
        for j in range(10):
            a[i].append(0)
    for i in range(10):
        a[i][0] = 1
        a[i][i] = 1
    for i in range(2,10):
        for j in range(1,i):
            a[i][j] = a[i - 1][j-1] + a[i - 1][j]
    from sys import stdout
    for i in range(10):
        for j in range(i + 1):
            stdout.write(a[i][j])
            stdout.write(' ')
        print
       
'''
题目62:学习putpixel画点。
1.程序分析:            
2.程序源代码:
#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);
}
'''
'''
题目63:画椭圆ellipse   
1.程序分析:
2.程序源代码:
'''
if __name__ == '__main__':
    from Tkinter import *
    x = 360
    y = 160
    top = y - 30
    bottom = y - 30
   
    canvas = Canvas(width = 400,height = 600,bg = 'white')
    for i in range(20):
        canvas.create_oval(250 - top,250 - bottom,250 + top,250 + bottom)
        top -= 5
        bottom += 5
    canvas.pack()
    mainloop()
'''
题目64:利用ellipse and rectangle 画图。
1.程序分析:
2.程序源代码:
'''
if __name__ == '__main__':
    from Tkinter import *
    canvas = Canvas(width = 400,height = 600,bg = 'white')
    left = 20
    right = 50
    top = 50
    num = 15
    for i in range(num):
        canvas.create_oval(250 - right,250 - left,250 + right,250 + left)
        canvas.create_oval(250 - 20,250 - top,250 + 20,250 + top)
        canvas.create_rectangle(20 - 2 * i,20 - 2 * i,10 * (i + 2),10 * ( i + 2))
        right += 5
        left += 5
        top += 10
    canvas.pack()
    mainloop()
       
'''
题目65:一个最优美的图案。   
1.程序分析:
2.程序源代码:
'''
import math
class PTS:
    def __init__(self):
        self.x = 0
        self.y = 0
points = []
def LineToDemo():
    from Tkinter import *
    screenx = 400
    screeny = 400
    canvas = Canvas(width = screenx,height = screeny,bg = 'white')
    AspectRatio = 0.85
    MAXPTS = 15
    h = screeny
    w = screenx
    xcenter = w / 2
    ycenter = h / 2
    radius = (h - 30) / (AspectRatio * 2) - 20
    step = 360 / MAXPTS
    angle = 0.0
    for i in range(MAXPTS):
        rads = angle * math.pi / 180.0
        p = PTS()
        p.x = xcenter + int(math.cos(rads) * radius)
        p.y = ycenter - int(math.sin(rads) * radius * AspectRatio)
        angle += step
        points.append(p)
    canvas.create_oval(xcenter - radius,ycenter - radius,
                       xcenter + radius,ycenter + radius)
    for i in range(MAXPTS):
        for j in range(i,MAXPTS):
            canvas.create_line(points[i].x,points[i].y,points[j].x,points[j].y)
    canvas.pack()
    mainloop()
if __name__ == '__main__':
    LineToDemo()
'''
题目66:输入3个数a,b,c,按大小顺序输出。   
1.程序分析:利用指针方法。
2.程序源代码:
'''
if __name__ == '__main__':
    n1 = int(raw_input('n1 = :n'))
    n2 = int(raw_input('n2 = :n'))
    n3 = int(raw_input('n3 = :n'))
    def swap(p1,p2):
        return p2,p1
    if n1 > n2 : n1,n2 = swap(n1,n2)
    if n1 > n3 : n1,n3 = swap(n1,n3)
    if n2 > n3 : n2,n3 = swap(n2,n3)
    print n1,n2,n3
'''
题目67: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。
1.程序分析:谭浩强的书中答案有问题。      
2.程序源代码:
'''
def inp(numbers):
    for i in range(9):
        numbers.append(int(raw_input('input a number:n')))
    numbers.append(int(raw_input('input a number:n')))
p = 0
def max_min(array):
    max = min = 0
    for i in range(1,len(array) - 1):
        p = i
        if array[p] > array[max] : max = p
        elif array[p] < array[min] : min = p
    k = max
    l = min
    array[0],array[l] = array[l],array[0]
    array[9],array[k] = array[k],array[9]
def outp(numbers):
    for i  in range(len(numbers)):
        print numbers[i]
if __name__ == '__main__':
    array = []
    inp(array)
    max_min(array)
    outp(array)
   
   
'''
题目68: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
1.程序分析:
2.程序源代码:
'''
if __name__ == '__main__':
    n = int(raw_input('the total number is:n'))
    m = int(raw_input('back m:n'))
    def move(array,n,m):
        array_end = array[n - 1]
        for i in range(n - 1,-1,- 1):
            array[i] = array[i - 1]
        array[0] = array_end
        m -= 1
        if m > 0:move(array,n,m)
       
    number = []
    for i in range(n):
        number.append(int(raw_input('input a number:n')))
    print 'orignal number:',number
    move(number,n,m)
    print 'after moved:',number
'''
题目69: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
   圈子,问最后留下的是原来第几号的那位。
1. 程序分析:
2.程序源代码:
'''
if __name__ == '__main__':
    nmax = 50
    n = int(raw_input('please input the total of numbers:'))
    num = []
    for i in range(n):
        num.append(i + 1)
    i = 0
    k = 0
    m = 0
    while m < n - 1:
        if num[i] != 0 : k += 1
        if k == 3:
            num[i] = 0
            k = 0
            m += 0
        i += 1
        if i == n : i = 0
    i = 0
    while num[i] == 0: i += 1
    print num[i]
'''
题目70: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。   
1.程序分析:
2.程序源代码
就这样吧
'''
if __name__ == '__main__':
    s = raw_input('please input a string:n')
    print 'the string has %d characters.' % len(s)
'''
题目71:编写input()和output()函数输入,输出5个学生的数据记录。
1.程序分析:
2.程序源代码:
使用list来模拟结构(不使用class)
stu = [string,string,list]
'''
N = 3
#stu
    # num : string
    # name : string
    # score[4]: list
student = []
for i in range(5):
    student.append(['','',[]])
def input_stu(stu):
    for i in range(N):
        stu[i][0] = raw_input('input student num:n')
        stu[i][1] = raw_input('input student name:n')
        for j in range(3):
            stu[i][2].append(int(raw_input('score:n')))
def output_stu(stu):
    for i in range(N):
        print '%-6s%-10s' % ( stu[i][0],stu[i][1] )
        for j in range(3):
            print '%-8d' % stu[i][2][j]
if __name__ == '__main__':
    input_stu(student)
    print student
    output_stu(student)
'''
题目72:创建一个链表。
1.程序分析:           
2.程序源代码:
'''
if __name__ == '__main__':
    ptr = []
    for i in range(5):
        num = int(raw_input('please input a number:n'))
        ptr.append(num)
    print ptr
'''
题目73:反向输出一个链表。   
1.程序分析:
2.程序源代码:
'''
if __name__ == '__main__':
    ptr = []
    for i in range(5):
        num = int(raw_input('please input a number:n'))
        ptr.append(num)
    print ptr
    ptr.reverse()
    print ptr
'''
题目74:连接两个链表。
1.程序分析:
2.程序源代码:
代码上好像只有,列表排序
'''
if __name__ == '__main__':
    arr1 = (3,12,8,9,11)
    ptr = list(arr1)
    print ptr
    ptr.sort()
    print ptr
'''
题目75:放松一下,算一道简单的题目。
1.程序分析:
2.程序源代码:
'''
if __name__ == '__main__':
    for i in range(5):
        n = 0
        if i != 1: n += 1
        if i == 3: n += 1
        if i == 4: n += 1
        if i != 4: n += 1
        if n == 3: print 64 + i
'''
题目76:编写一个函数,输入n为偶数时,调用函数求1/2+1/4+...+1/n,当输入n为奇数时,调用函数
   1/1+1/3+...+1/n(利用指针函数)
1.程序分析:
2.程序源代码:
'''
def peven(n):
    i = 0
    s = 0.0
    for i in range(2,n + 1,2):
        s += 1.0 / i
    return s
def podd(n):
    s = 0.0
    for i in range(1, n + 1,2):
        s += 1 / i
    return s
def dcall(fp,n):
    s = fp(n)
    return s
if __name__ == '__main__':
    n = int(raw_input('input a number:n'))
    if n % 2 == 0:
        sum = dcall(peven,n)
    else:
        sum = dcall(podd,n)
    print sum
'''
题目77: 9*9 乘法口诀
# -*- coding: cp936 -*-
#enconding: utf-8
def gen(line_cnt):
    ''':
     @param:行数
    '''
    for i in range(1,line_cnt+1):
        for j in range(1,i+1):
            m=i*j
            print '%s*%s=%st' % (i,j,m),#这边的逗号很重要,有了逗号,才能不换行
        print''
 
if __name__ == '__main__':
    gen(9)
   
   
    #求可用被17整除的所有三位数
for num in range(99,1000):
    if num % 17 == 0:
        print num
题目78:写一个程序,提示输入整数X,然后计算从1到X连续整数的和
sum = 0
x = int (raw_input ('pls input x: '))
#输入函数
for num in range(0,x):
     sum += num
print sum
 
题目79:改写上面的计算过程,循环嵌套计算连续的和
x = int(raw_input('please input x : ' ))
layer = 1
while layer <= x:   
    lsum = 0
    count =1
    while count <=layer:
        lsum += count
        count += 1
        fstr = ''
        for num in range(1 , layer ):
           
            fstr = fstr + str(num) + '+'
        fstr += str(layer)
        print fstr,'=',lsum
    layer += 1
            # 如何控制空格的数量让等号在一起呢
           
题目80:有1、2、3、4个数字,能组成多少个互不相同且无重复数字的三位数?都是多少?
cnt = 0#count the sum of result
for i in range(1,5):
    for j in range(1,5):
        for k in range(1,5):
            if i!=j and i!=k and j!=k:
                print i*100+j*10+k
                cnt+=1
print cnt

"""

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