解法1cinclude<stdio.hintmain(){intnum=0;intcount=0;scanf("%d",&num);while(num){if(1num%2)count;num=num/2;}printf("二进制中1的个数=%d\n",count);return0;}//如果是负数，可以试试1，只循环一次（此代码存在问题）解法2(按位与法)cinclude<stdio.hintm
解法1（引第三变量）cinclude<stdio.hintmain(){inta=3;intb=5;inttmp=0;printf("before：a=%db=%d\n",a,b);tmp=a;a=b;b=tmp;printf("after：a=%db=%d\n",a,b);return0;}//效率高解法2（加减法）cinclude<stdio.hintmain(){inta=3;intb=5;
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