【10】选（B）这个……A项少了分号，C项既然是指针，那么取的应该是地址，最后一项，显然……字符串，如果不是GETS（）函数，那么就应该用循环，才能实现SCANF的输入 【11】选（B）显然example是结构体内某个具体的对象，而x，y，z都是关于example属性修饰之类的成员，比如说example可以代表一本例子...
{ count = 0;break;} n[count] = str[i]-'a'+1;//字符串转换为数字量并存储 count ++;c = str[i];} } if (0 == count){ cout<<count<<endl;} else { for (int j=0;j<count;j++){ cout<<n[j]<<" ";} } cout<<endl;} main...
s;}while(number) {ch = number%k;if(ch > 9) s[n] = ch - 10 + 'A';else s[n] = ch + '0';number /= k;++n;}s[n] = '\0';for(i = 0; i < n/2; ++i) {ch = s[i];s[i] = s[n - i - 1];s[n - i - 1] = ch;}return s;}int main() {...
; break; } //T字顺转180度case 7: { a[x][y-1]=b[1]; a[x][y+1]=b[2]; a[x+2][y]=b[3]; break; } //T字顺转270度case 8: { a[x][y+1]=b[1]; a[x-2][y]=b[2]; a[x+2][y+1]=b[3]; break; } //Z字方块case 9: { a[x][y-1]=b[...
int i,j=0;for(i=0;i<N;i++)if(c[i]>'a'&&c[i]<'z') j++;return j;}void main(){int i,j;char a[N];printf("请输入10个字符:");for(i=0;i<N;i++)scanf("%c",&a[i]);j=b(a);printf("小写字母的个数为:%d",j);}第七题：include<stdio.h>include<string....
C. scanf("%d%d%d",x,y,z); D. read("%d%d%d",&x,&y,&z);17.以下数组定义中不正确的是（）A) int a[2][3]; B) int b[][3]={0,1,2,3};C) int c[100][100]={0}; D) int d[3][]={{1,2},{1,2,3},{1,2,3,4}};18. 以下程序的输出结果是（）main()...
include<stdio.h> include<math.h> int max(int x,int y){ if(x>y)return x;else return y;} int min(int x,int y){ if(x<y)return x;else return y;} int main(){ int x,y,a,b,c,d,e,f,g,h,i,t,o,p,l;printf("输入任意五个整数:\n");scanf("%d%d%d",&a,&b...
include <stdio.h>void main( ){ char s[20], ch; int i,j; printf("Please input a string:\n"); gets(s);//输入字符串 printf("Please input acharacter:\n"); ch=getchar();//输入一个字符。 for(i=0;i<20;i++)//对于每个位置做遍历。 { fo...
break; default:printf("error\n"); } return 0;}2include<stdio.h>main(){ int i,j; printf("1-5000之间能同时被3、5、7除可以除尽的有\n"); for(i=1,j=0;i<=5000;i++) { if((i%3==0)&&(i%5==0)&&(i%7==0)) { printf("%d ",i); j++; ...
c as IntegerFor a = 1 To 10000c = 0For b = 1 To a \ 2If a Mod b = 0 Then c = c + bNext bIf a = c Then Print Str(a)Next a5.利用C语言编程求1000以内完全数#include stdio.hvoid main(){int j,k,sum = 0;for(k=2;k<=1000;k++){sum=0;for(j=1;j<k;j...