![有以下程序 main() { int p[8]={11,12,13,14,15,16,17,18},i=0,j=0; while(i++<7) if(p[i]%2),第1张 有以下程序 main() { int p[8]={11,12,13,14,15,16,17,18},i=0,j=0; while(i++<7) if(p[i]%2),第1张](/aiimages/%E6%9C%89%E4%BB%A5%E4%B8%8B%E7%A8%8B%E5%BA%8F+main%28%29+%E3%80%80%E3%80%80%7B+%E3%80%80%E3%80%80int+p%5B8%5D%3D%7B11%2C12%2C13%2C14%2C15%2C16%2C17%2C18%7D%2Ci%3D0%2Cj%3D0%3B+%E3%80%80%E3%80%80while%28i%2B%2B%26amp%3Blt%3B7%29+if%28p%5Bi%5D%252%29.png)
if(p[i]%2) 等同 if(p[i]%2==1)
i = 0, <7, i++ 得1,p[i]=12, p[i]%2=0 为假
i = 1, <7, i++ 得2,p[i]=13, p[i]%2=0 为 真 j=13
i = 2 <7, i++ 得3,p[i]=14, p[i]%2=0 为假
i = 3, <7, i++ 得4,p[i]=15, p[i]%2=0 为 真 j=13+15
i = 4, ,,,为假
i = 5, <7, i++ 得6,p[i]=17, p[i]%2=0 为 真 j=13+15+17
i= 6, ,,,为假
最终 j=13+15+17=45
应该选D24*(p+i)%2是计算元素的奇偶,11和13是奇数,*(p+i)%2结果值均为1,且i<7,但14是偶数,*(p+i)%2结果值为0,循环条件不成立退出循环,所以k是11+13=24
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