设证书i=1234,编写程序,要求i的值变为4321

#include#includemain(){longsum=1;inti;intsum0=0;clrscr();for(i=1;i

楼上两位的代码可能有点小问题s += m*m+1/m;改成s += m*m+1.0/m;再问：您好，请您帮忙写一个完整的编程好么？谢谢了再答：#include&

#includemain(){floatx=2.5,a=7,y=4.7;printf("%f",x+a%3*(x+y)%2/4);}#includemain(){inta=2,b=3;floatx=3

MOVEAX,AMOVEBX,BMOVECX,CADDEAX,EBXSUBEAX,ECXMOVY,EAX再问：运行不了，好多错误啊再答：我假设你声明了A,B,C,Y这几个变量。。具体问题有疑问可以私密

#includevoidmain(){intx=0,y;printf("输入X\n");scanf("%d",&x);if(x=1&&x=10){y=3*x-11;}printf("y=%d\n",y

#includevoidmain(){inta=7;floatx=2.5,y=4.7,r;r=x+a%3*int(x+y)%2/4;printf("%f\n",r);}运行结果2.500000

*程序一*/main(){intn,i=2,isPrime=1;printf("Inputn:\n");scanf("%d",&n);while(i

%创建m文件文件名为f%编写函数function[y]=f(x)ifx6y=0.5;elsey=1.5-0.25*x;end%画图x=0:0.01:2;plot(x,f(x));

varn,i,min,max,maxi,mini,s:integer;x:array[1..100]ofinteger;ans:real;beginreadln(n);fori:=1tondoread

pt = {2, 2};ContourPlot[ Sqrt[(x - pt[[1]])^2 + (y -&nb

start:movax,30addax,10subax,77movdl,axmovah,02hint21hmovah,07hint21hmovah,4chint21hendstartstart：mov

#include"stdafx.h"voidmain(){ints=0;for(intn=1;n

publicclassTest{publicstaticvoidmain(String[]args){ints=0;intn=1;for(inti=0;i

/>vart;varl=prompt('请输入边长',3.5);t=3.1415926*(l/2.0)*(l/2.0);document.write("边长:"+l);document.write("

;MOVAX,AANDAX,B;AX=aANDbMOVBX,AXORBX,B;BX=aXORbADDAX,BXADDAX,BX;AX=2*(aXORb)+aANDbADDAX,A;AX=a+2*(aX

CLEARSETTALKOffs1=0forn=1to10s1=s1+jc(n)next"1!+2!+3!+.+10!=",s1FUNCTIONjcPARAMETERSis=1FORj=1TOis=s

int.Parse(i);或者(int)(i);

[x,y,z]=ellipsoid(0,0,1,2,3,4);t=-1:0.1:1;[X,Y,Z]=cylinder(1+t.^2);subplot(2,2,1);surfl(x,y,z);subpl

//看看这个怎么样希望采纳哦!// Note:循环、迭代求 1 + 12 + 123 + 1234 +...+

/*输入参数a,b,c,使用空格隔开*/#include#include#includeintmain(){inta,b,c;floatx1,x2;printf("Inputa,b,c:\n")