x = 0for i in range(3): x+=1 for j in range(3): if(j): continue x+=1print(x)
6
举一反三
- x = 0for i in range(3): x+=1 for j in range(3): if(j): continue x+=1print(x)输出结果为（ ）
- 执行以下程序之后x的值为（）x=0for i in range(3):x+=1for j in range(3):if(j):continuex+=1print(x) A: 3 B: 6 C: 9 D: 12
- 执行以下程序后，x的值是x=0for i in range(3):x+=1for j in range(3):if(j):continuex+=1 A: 12 B: 9 C: 6 D: 3
- 执行以下程序后，x的值是多少 x = 0 for i in range(3): x += 1 for j in range(3): if (j): continue x += 1
- for i in range(1, 3): for j in range(1, 3): if i * j <; 2: continue print i * j
内容
- 0
以下程序的输出结果是__________。 x= 0 while x<6: if x%2==0: continue if x==4: break x+=1 print("x=",x)
- 1
a = [x for x in range(4) if x % 2 ==1]，语句print(a)输出为 A: [1, 2, 3] B: [0, 1, 2, 3] C: [0, 2] D: [1, 3]
- 2
关于下面代码的叙述，正确的是哪一项？ x=0 while x<10: x+=1 print(x) if x>3: break
- 3
方程${{x}^{2}}{{y}^{''}}-(x+2)(x{{y}^{'}}-y)={{x}^{4}}$的通解是( ) A: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$ B: $y={{C}_{1}}x+{{C}_{2}}{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ C: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{4}})$ D: $y={{C}_{1}}x+{{C}_{2}}x{{e}^{x}}-(\frac{1}{2}{{x}^{3}}+{{x}^{2}})$
- 4
main() { int i,j,x=0; for(i=0;i<2;i++) { x++; for(j=0;j<=3;j++) { if(j%2) continue; x++; } x++; } printf(“x=%d ”,x); }