请帮我看一下这个判断程序intx,y,z,oddcount=0;printf("Input x,y,z:");scanf("%d,%d,%d",&x,&y,&z);x%2==1?oddcount++:0;y%2==1?oddcount++:0;z%2==1?oddcount++:0;oddcount==2?printf("YES"):printf("NOT");在三目运算当中’‘x%2==1’‘,它
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![请帮我看一下这个判断程序intx,y,z,oddcount=0;printf(](/uploads/image/z/14505690-66-0.jpg?t=%E8%AF%B7%E5%B8%AE%E6%88%91%E7%9C%8B%E4%B8%80%E4%B8%8B%E8%BF%99%E4%B8%AA%E5%88%A4%E6%96%AD%E7%A8%8B%E5%BA%8Fintx%2Cy%2Cz%2Coddcount%3D0%3Bprintf%28%22Input+x%2Cy%2Cz%3A%22%29%3Bscanf%28%22%25d%2C%25d%2C%25d%22%2C%26x%2C%26y%2C%26z%29%3Bx%252%3D%3D1%3Foddcount%2B%2B%3A0%3By%252%3D%3D1%3Foddcount%2B%2B%3A0%3Bz%252%3D%3D1%3Foddcount%2B%2B%3A0%3Boddcount%3D%3D2%3Fprintf%28%22YES%22%29%3Aprintf%28%22NOT%22%29%3B%E5%9C%A8%E4%B8%89%E7%9B%AE%E8%BF%90%E7%AE%97%E5%BD%93%E4%B8%AD%E2%80%99%E2%80%98x%252%3D%3D1%E2%80%99%E2%80%98%2C%E5%AE%83)
请帮我看一下这个判断程序intx,y,z,oddcount=0;printf("Input x,y,z:");scanf("%d,%d,%d",&x,&y,&z);x%2==1?oddcount++:0;y%2==1?oddcount++:0;z%2==1?oddcount++:0;oddcount==2?printf("YES"):printf("NOT");在三目运算当中’‘x%2==1’‘,它
请帮我看一下这个判断程序
intx,y,z,oddcount=0;
printf("Input x,y,z:");
scanf("%d,%d,%d",&x,&y,&z);
x%2==1?oddcount++:0;
y%2==1?oddcount++:0;
z%2==1?oddcount++:0;
oddcount==2?printf("YES"):printf("NOT");
在三目运算当中’‘x%2==1’‘,它是要理解成‘’x能被2整除是真‘’么,它和‘’x%2==0‘’能整除是不是矛盾了
请帮我看一下这个判断程序intx,y,z,oddcount=0;printf("Input x,y,z:");scanf("%d,%d,%d",&x,&y,&z);x%2==1?oddcount++:0;y%2==1?oddcount++:0;z%2==1?oddcount++:0;oddcount==2?printf("YES"):printf("NOT");在三目运算当中’‘x%2==1’‘,它
’‘x%2==1’‘,它是要理解成‘’x除以2的余数是1‘’,即x是一个奇数.