1 条题解

  • 0
    @ 2023-6-11 12:17:11

    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

    信息

    ID
    2383
    时间
    1000ms
    内存
    32MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者