1.编写一个程序,计算5*5矩阵中主对角线元素之和.

#includevoidmain(){floatsum=0.0;intn=0,sign=-1;scanf("%d",&n);for(floati=1;i

privatesubform_click()dimnaslong,snaslongn=inputbox("请输入n")fori=0tonsn=sn+2*i+1nextmsgbox"1+3+5+...(

1.#include <stdio.h>int main(){    float Income,Tax; &

//别说100位,1000位都没问题,给你个例子.#include#definePRECISION2800#defineFRACTION1000#defineGROUP4#defineINITIALV

PrivateSubCommand1_Click()Dima,xAsSinglea=Val(Text1.Text)Ifa再问：x=x+a*0.2+3000*0.2怎么都乘0.2呀？

INPUT"a=",aINPUT"b=",bPRINT"|a|+|b|=";abs(a)+abs(b)#includeinta,b;scanf("%d",&a);scanf("%d",&b);prin

VB程序PrivateSubCommand1_Click()Dima,iAsIntegeri=1Whilei<50s=s+(i+1)/ii=i+1WendPrint"s=";sEndSu

#includeintfact(intn){if(n==0||n==1)return1;elsereturnn*fact(n-1);}intmain(){printf("5!+10!=%d\n"

#includevoidmain(){floata,b,result;charc;printf("pleaseinputastatement:");scanf("%f%c%f",&a,&c,&b);s

#include#includevoidmain(){intsum,i;sum=0;for(i=1;i

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

程序框图如下：程序如下：s=1i=3DOs=s*ii=i+2LOOPUNTILi＞99PRINTsEND

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

#include <stdio.h>int main(){  float x, tax = 0; 

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

publicclassTestArray{publicstaticvoidmain(String[]args){intm=0;for(inti=1;iintn=1;for(intj=1;jn=n*j;

/>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("

doublefunction(intn){doublevalue=0;for(inti=1;i

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

#includeintmain(){intsum=0;inti=1;intj=1;for(;i