编写程序,是给定的一个5*6的二维整形数组转置
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///c语言的,#inlcudevoidmain(){inta;printf("pleaseinputanumber:\n");scanf("%d",&a);if(a%5)printf("thisnu
ModuleModule1SubMain()DimA(,)AsInteger={{1,2,3,4,5,6,7,8,9,0},{1,2,3,4,5,6,7,8,9,0},{1,2,3,4,5,6,7,8
voidmain(){intn,i,s;s=1;printf("pleaseinputn:");scanf("%d",&n);for(i=2;i
判断n是否为素数,可用n除以2到根号n之间所有的整数,除不尽则n为素数Fora=2ToInt(Sqr(n))IfnModa0Then'余数不为0,则n为素数isprime=True'n为素数Elsei
这个东西,我一时也写不出来,它主要的思想是中缀表达式转成后缀表达式,然后后缀表达式求值,这两部都需要堆栈处理.
源程序如下:#includeintmain(){doubley,x,n,m;printf("y=");scanf("%lf",&y);printf("请输入x的取值区间:\n");scanf("%lf
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?
#includeclassangle{private:floatx,y,z;public:angle();voidthreeb(floata,floatb,floatc);friendvoidcoma
while(a){case‘1‘intn=rand()%10;break;case‘1‘intn=10+rand()%100;break;case‘1‘intn=100+rand()%1000;bre
includeincludeddoublefun(inta,intb,intc){intp;p=(a+b+c)/2;returnsqrt(p*(p-a)*(p-b)*(p-c));}再问:ok再问:
inttemp1;intlength=0;intlength2=0;for(inti=0;i{if(i==0)temp=b[i];else{if(temp==b[i])length++;else{if
#include#include#includechar*getline()//读取一行,动态分配内存,此函数在网上搜到的,遇到问题先搜搜:){char*line=malloc(100),*linep
#includevoidmain(){doublea=0,b=0;printf("请输入矩形的长:");scanf("%lf",&a);printf("请输入矩形的宽:");scanf("%lf",&
这一行:DoWhilex>a(p)Andp
#include <stdio.h>char* dg(char* instr, char* outstr, char* 
程序框图如下:程序如下:s=1i=3DOs=s*ii=i+2LOOPUNTILi>99PRINTsEND
importjava.util.*;publicclassShuShu{publicbooleanisShuShu(intn){inti;for(i=2;i=n){returntrue;}else{r
用VC6.0帮你现写的:#include <stdio.h>void zhihuan(int (*p)[4]);int main(){ &
/>import java.util.Scanner;public class MyNum {\x09\x09private static
"计算两个给定的长方形的周长和面积"是两个长方形还是一个长方形..需要输入什么条件..如果是输入长,宽,然后计算面积的话..那么#include"stdio.h"voidmain(){floatcha