1 solutions
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0
C :
#include<stdio.h> #include<stdlib.h> int main(){ double eletric,money; scanf("%lf",&eletric); if(eletric<=150) money=eletric*0.4463;//电量少于150KWh时的电费 if(eletric>150&&eletric<=400) money=150*0.4463+(eletric-150)*0.4663;//电量多于150KWh,少于400KWh时的电费 if(eletric>400) money=150*0.4463+250*0.4663+(eletric-400)*0.5663;//电量多于400KWh时的电费 printf("%.1lf",money);//保留小数点1位 printf("\n");//换行 return 0; }
Python :
n = int(input()) if n <= 150: a = n * 0.4463 print(round(a,1)) if n > 150 and n <= 400: a = 150 * 0.4463 + (n - 150) * 0.4663 print(round(a,1)) if n > 400: a = 150 * 0.4463 + 250 * 0.4663 + (n - 400) * 0.5663 print(round(a,1))
- 1
Information
- ID
- 10459
- Time
- 1000ms
- Memory
- 32MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By