编程.计算并输出x(包括x)以内所有能被3或7整除的自然数的倒数之和

#include<stdio.h>#include<math.h>void main(){float x,y;scanf("%f",&a

#includeintmain(){\x09intn,sum=0;\x09scanf("%d",&n);\x09while(n)\x09{\x09\x09sum+=n%10;\x09\x09n/=10

publicclassTest{publicstaticvoidmain(String[]args){System.out.println(show());}publicstaticdoublesho

#include#includeusingnamespacestd;voidmain(){doublea[10];inti;for(i=1;i

longfunc(intm){inti,sum=1;for(i=1;i

#includevoidmain(){intx,y;scanf("%d",&x);if(x>10)y=2*x*x-x+1;elseif(x>2)y=3*x+6;elsey=0;printf("%d",

第一题#includevoidmain(){intx,y,z;printf("请分别输入x,y,按Enter结束\n");scanf("%d",&x);scanf("%d",&y);z=(x

#includeintmain(void){doublex;scanf("%lf",&x);if(x{printf("y=0\n");}elseif(x>=0.0&&x{printf("y=%lf\n

基于c#的classProgram{staticvoidMain(string[]args){Console.WriteLine("Pleaseenteranumber");inta=int.Pars

#includeintmain(void){intx,y;y=x*x;printf("%d=%d*%d",y,x,x);printf("%d*%d=%d",x,x,y);return0;}再问：运行结

#include#includedoubleab(doublen){inti;doubles=1;for(i=1;ipow(10,-5)){c=pow(-1,j-1)*pow(x,j)/ab(j);d

#include#include#includevoidmain(){doublex,y,temp;cin>>x;if(x==0){exit(0);cout

定义变量x,y;输入x;if(xy=fabs(2*x+5);}elseif(x=0){y=sqrt(x+1);}else{y=3*x+1;}输出y;return1;大体就是这样,因为我不是学c语言的,

vari,x,s:longint;beginfori:=1to100dobeginx:=2+i*10;s:=s+x;end;writeln(s);end.

#include//希望对您有用#includeintmain(){intx,y;printf("Pleaseenterthenumber:");scanf("%d",&x);if(x

#include#includetypedefstructPoint{doublex,y;}Point;voidmain(){Pointp1,p2;doublek,b;printf("输入点

voidmain(){floatx,y;cin>>x;if(x=10)y=3*x-11;elsey=2*x-1;cout

oot=sqrt(x);

voidmain()intx,abs;printf("shurux");scanf("%d",&x);if(x>0)abs=x;elseabs=0-x;printf("jueduizhishi%d",

#includeusingnamespacestd;intmain(){\x09doublex,y;\x09cout