用switch语句编写函数,输入一个小于六位数的整数,判断它是几位并逆袭输出

n=input('请输入成绩：');k=floor(n/10);switchkcase{8,9,10}disp('优秀')case7disp('良好')case6disp('一般')otherwise

main(){inta,b,c;floats;printf("pleaseentera:\n");scanf("%d",&a);b=a-1200;if(b>0&&b500&&b2000&&b5000&

clearinput"请输入x的值："toxdocasecasex=2andx10y=-3*xendcase"y的值是：",y

//if语句#inlcude#includedefinea10//a自己取值voidmain(){floatx,y;scanf("%f",&x);if(x>=0.5&&x=1.5&&xy=exp(a*

#includeintmain(void){intscore;chargrade;printf("enterthetestscore(0-100):");scanf("%d",&score);if(s

#includevoidmain(){intx;inty;printf("请输入X的值,-5再问：请问用嵌套语句怎么写,谢谢谢谢再答：可以不用嵌套就写出来干嘛那么麻烦再问：呵呵,一个考试题目,想搞明白

#includevoidmain(){intx,k;scanf("%d",&x);if(x-5)k=-1;elseif(x=0)k=0;elseif(x>0&&x

#includemain(){intn;floatprice,money,r;intdiscount;//折扣printf("请输入商品购买量和单价:\n");scanf("%d,%f",&n,&pr

请参照课本自己写,这种东西很简单的再问：老师不让抄啊再答：inti=int.Parse(Console.ReadLine());switch(i){case1:Console.WriteLine("E

if(x>=-5&&x=0&&x=5&&x=10&&x=-5&&x=0&&x=5&&x=10&&x

#includemain(){intx,y;printf("pleaseenterdatax\n);scanf("x=%d\n",&x);switch(y){case(x=10&&x=20&&x

floatx;intc;c=(int)(x+0.5);//四舍五入switch(c)case1:a+bxcase2:a-bxcase3:a*bxcase4:a/(bx)大概思路就是这样,最好你自己写程

你好很高兴为你假设每件上衣和裤子都是成套的GetPrice(intshirtCount,intpantsCount)intsuiteCount=0;intminus=shirtCount-pantsC

#includevoidmain(){floatnumber,sum;printf("pleaseinputanumber:");scanf("%f",&number);if(number100000

switch(s){case1:x=1;break;case2:x=3;break;case3:case4:case5:case6:x=2;break;case7:case8:x=1;break;ca

case后面只能跟一个固定的整数不能跟范围你可以在switch前面加一句intj；j=i/10;然后根据J的值case还有程序要求bonus是长整形的改一下

main(){inta,b,c;//a是输入测试值,b是分段值,c是条件标志printf("inputthevalueandpercentils:\n");scanf("%d%d",&a,&b);//

y=1./(1+25*x.*x)

publicintswitchTest(intx){intflag=0;if(x-5){flag=1;}elseif(x==0){flag=2;}elseif(x>0&&xflag=3;}switch

privatedoubleJJ(doubles,doublee,doublep,doublec){returnce?e:c)*p}doubleJ=奖金大大的JJJ=JJ(0,10,0.1,J)+JJ(