编写一个程序,给定一个20个元素以内的整数数组,求数组中最大元素的值及下标
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#include#include#include#includeusingnamespacestd;structWord{charword[30];//存放单词//单词总数就是结构体变量数组大小int
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
#include<stdio.h>main(){ int a[]={0,1,2,3,4,5,6,7,8,9},i; &nb
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
1.思路:用一个字符串数组储存输入的数据,并分别输出。程序:Vara:array[1..3]ofstring;i:integer;BeginFori:=1to3doReadln(a[i]);Fori:
#includeclassangle{private:floatx,y,z;public:angle();voidthreeb(floata,floatb,floatc);friendvoidcoma
programa4;vara,b,c:integer;beginreadln(a,b,c);if(((a>0)and(b>0)and(c>0))and((a+b>c)and(a+c>b)and(b+c
inttemp1;intlength=0;intlength2=0;for(inti=0;i{if(i==0)temp=b[i];else{if(temp==b[i])length++;else{if
这个可以这样做~cleara=rand(10,1);%产生一组随机数b=rand(10,1);%产生另一组随机数b=b-(dot(a,b)/dot(a,a)).*a;%可以使用施密特正交化的方法转化d
namespaceConsoleApplication1{classTest{publicstaticvoidMain(){inti,j,k,n=100;for(i=0;i
#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
varn,i,min,max,maxi,mini,s:integer;x:array[1..100]ofinteger;ans:real;beginreadln(n);fori:=1tondoread
#include <stdio.h>char* dg(char* instr, char* outstr, char* 
pt = {2, 2};ContourPlot[ Sqrt[(x - pt[[1]])^2 + (y -&nb
importjava.util.*;publicclassShuShu{publicbooleanisShuShu(intn){inti;for(i=2;i=n){returntrue;}else{r
/>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("
/>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