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【入门2】分支结构 - 题单 - 洛谷https://www.luogu.com.cn/training/101#problems一、【深基3.例2】数的性质 - 洛谷
一些数字可能拥有以下的性质:
小A 喜欢这两个性质同时成立的数字;Uim 喜欢这至少符合其中一种性质的数字;八尾勇喜欢刚好有符合其中一个性质的数字;正妹喜欢不符合这两个性质的数字。
一、暴力写法
- num = int(input())
-
- if num >=0 and num <= 1000:
- if (num % 2 == 0) and (num > 4 and num <= 12):
- print("1",end=" ")
- else:
- print("0",end=" ")
- if (num % 2 == 0) or (num > 4 and num <= 12):
- print("1",end=" ")
- else:
- print("0",end=" ")
-
- if (num % 2 != 0) and (num > 4 and num <= 12):
- print("1",end=" ")
- elif (num % 2 == 0) and (num <= 4 and num > 12):
- print("1",end=" ")
- else:
- print("0",end=" ")
-
- if (num % 2 == 0) and (num > 4 and num <= 12):
- print("0",end=" ")
- else:
- print("1",end=" ")
二、Python选择结构特有写法
- n = eval(input())
- p1 = n % 2 == 0
- p2 = 12 >= n > 4
- print(1 if p1 and p2 else 0,end=" ")
- print(1 if p1 or p2 else 0,end=" ")
- print(1 if (p1 and not p2) or (not p1 and p2) else 0,end=" ")
- print(1 if not p1 and not p2 else 0,end="")
分析:选择结构的两种写法
1. 表达式为真输出区 if 表达式 else 表达式为假输出区
2.[表达式为假输出区,表达式为真输出区][判断表达式]
输入一个年份(大于 1582 的整数 ),判断这一年是否是闰年,如果是输出 1,否则输出 0。
方法一:算法实现
- year = int(input())
- if year > 1582:
- if year % 4 ==0 and year %100 != 0 or year % 400 == 0:
- print(1)
- else:
- print(0)
方法二:调用库函数
- year = int(input())
- if year > 1582:
- if isleap(year):
- print(1)
- else:
- print(0)
分析:简单的闰平年判断题
- num = int(input())
-
- if 0<=num<=100:
- if num<=1:
- print("Today, I ate {} apple.".format(num))
- else :
- print("Today, I ate {} apples.".format(num))
- num = int(input())
-
- Local = num*5
-
- Luogu = num*3+11
-
- if num<=100:
- if Local < Luogu:
- print("Local")
- elif Local == Luogu:
- print("Local"+"和"+"Luogu"+"时间一样")
- else:
- print("Luogu")
- m,h = map(float,input().split())
- if 40<=m<=120 and 1.4<=h<=2:
- BMI = m/(h*h)
- if BMI < 18.5:
- print("Underweight")
- elif 18.5 <= BMI < 24:
- print("Normal")
- elif BMI >= 24:
- print("%g"%BMI)
- print("Overweight")
- a,b,c = map(int,input().split())
- p = 0<= a,b,c <=100
- if p:
- if a>=b:
- a,b =b,a
- if a>=c:
- a,c=c,a
- if b>=c:
- b,c=c,b
- print("{} {} {}".format(a,b,c))
- y, m = map(eval, input().split())
- if m ==2 :
- p1 = y % 4 == 0 # 被4整除是闰
- p2 = y % 100 == 0 # 被100整除不是闰
- p3 = y % 400 == 0 # 是闰
- if p3 or (p1 & (not p2)):
- print(29)
- else:
- print(28)
- else:
- if m in (1, 3, 5, 7, 8, 10, 12):
- print(31)
- else:
- print(30)
- max_cd = 0
- day = 0
- for i in range(1,8):
- a,b = map(int,input().split())
- if (a+b)-8 > max_cd:
- max_cd = a+b -8
- day = i
- print(day)
- n_need=int(input())
- money = [None]*3
- for kind in range(3):
- num,PenMoney = map(int,input().split())
- for i in range(1,100000):
- if num*i>=n_need:
- money[kind]=i*PenMoney
- break
- print(min(money))
- ISBN = input()
- strNum = ISBN[:12]
- Sum = 0
- j = 1
- for i in strNum:
- if i != '-':
- n = int(i)
- Sum = Sum + n * j
- j += 1
- else:
- continue
- if Sum % 11 == 10:
- ID = 'X'
- else:
- ID = str(Sum % 11)
- if ID == ISBN[12]:
- print("Right")
- else:
- strNum = strNum + ID
- print(strNum)
十一、小玉家的电费 - 洛谷
- n = int(input())
- s = 0.0
- if n <= 150:
- s = 0.4463 * n
- elif n <= 400:
- s = (n - 150) * 0.4663 + 150 * 0.4463
- else:
- s = (n - 400) * 0.5663 + 250 * 0.4663 + 150 * 0.4463
- print('%.1f' % s)
- x, n = map(int, input().split())
- ans = (n//7)*5
- n %= 7
- while n > 0 :
- if x != 6 and x != 7 :
- ans += 1
- x += 1
- n -= 1
-
- print(ans*250)
十三、三角函数 - 洛谷
- from fractions import Fraction
- list_1 = list(map(int,input().split()))
- list_1 = sorted(list_1)
- a1,a3 = list_1[0],list_1[2]
- a = Fraction(a1, a3)
- print(a)
- a = input().split(' ')
- a = [int(i) for i in a]
- a = sorted(a)
- a1 = a[0]
- a3 = a[2]
- # 约分
- while a1:
- yushu = a3%a1
- a3 = a1
- a1 = yushu
- print ("%d/%d"%(a[0]/a3,a[2]/a3))
- AppHeight = map(int,input().split())
- TTHeight = int(input())
- count = 0
- if 100<TTHeight<120:
- for item in AppHeight:
- if 100<=item<=200:
- if TTHeight+30>=item:
- count+=1
- print(count)
- #include<stdio.h>
- int main(){
- int a,b,c; //三边
- scanf("%d%d%d",&a,&b,&c);
- int t; //临时
- //先由小到大排列三边 以便实现三角形的特性
- if(a>b){
- t=a;
- a=b;
- b=t;
- }
- if(a>c){
- t=a;
- a=c;
- c=t;
- } if(b>c){
- t=b;
- b=c;
- c=t;
- }
- if(a+b<=c){ //三角形 任意两边要大于第三边 按大小排序过 所以a+b>c就是三角形
- printf("Not triangle");
- }else{ //一定把判断三角形当作一个块 其他的全放到else里 因为1 3 3不是三角形但是它满足等腰三角形的要求
- if(a*a+b*b==c*c){
- printf("Right triangle\n");
- }
- if(a*a+b*b>c*c){
- printf("Acute triangle\n");
- }
- if(a*a+b*b<c*c){
- printf("Obtuse triangle\n");
- }
- if(a==b||b==c){ //锐角等腰 两长边一短边 钝角等腰 一长边两短边
- printf("Isosceles triangle\n");
- }
- if(a==b&&b==c){
- printf("Equilateral triangle");
- }
- }
- return 0;
- }
- a,b,c = map(int,input().split())
- order = input().upper()
- if a+b+c<=300:
- MinNum =min(min(a,b),c)
- MaxNum =max(max(a,b),c)
- MidNum = (a+b+c)-MinNum-MaxNum
- for item in order:
- if item == 'A':
- print(MinNum,end=" ")
- if item == 'B':
- print(MidNum,end=" ")
- if item == 'C':
- print(MaxNum,end=" ")
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