- 首页
- 全科主治
答案:
-
1.0=x00=0xa I=A0≠2
-
2.有以下程序 #include main() { int a[ ]={2,3,5,4},i; for(i=0;i switch(i%2) { case 0:switch(a[i]%2) {
-
3.判断下列代码段的大O级别i = nwhile i > 0: k = 2 + 2 i = i // 2i = nwhile i > 0: k = 2 + 2 i = i // 2-
-
4.for i in range(1,5):if i%2==0:print(i)break
-
5.A.for (i=1; ; ) { if (i %2==0) continue ; if (i %3==0) break ; }B.i=32767; do { if (i
-
6.y = 0 for i in range(0, 10, 2): y = i print(y)A.30B.20C.10D.9
-
7.下面程序的运行结果为 void main(){ int a[2][3]={5,6,4,7,8,9}; int m[2],i,j; for(i=0;i<2;i++) {
-
8.2)k=0;for (i=1: i=n: i Or
-
9.有如下Visual Basic程序段: s = 0 For i = 10 To 1 Step -1 If i Mod 2 = 0 And i Mod 5 = 0 Then s = s + i Next
-
10.#define A 100 main() { int i=0,sum=0; do{ if(i==(i/2)*2) continue; sum+=i; }while(++i<A) ; printf("%d