2876. 二进制位不同的个数

include

    using namespace std;
    bitset<100>grath;
    int main()
    {
            int t;cin>>t;
            while(t--){
                    int a,b;
                    cin>>a>>b;
                    grath=a^b;
                    cout<<grath.count()<<endl;
            }
    }

我觉得有个要注意的地方，如果输入的x和y位数长短不一致，比如x比y位数长。y的高位要补0再和x比，x的高位中的1视作和y不一致的位。

瑞哥又来送答案啦

include

include

int main()
{
int t,x,y,z,q;
scanf(“%d”,&t);
while(t–)
{
scanf(“%d %d”,&x,&y);
z=0;
for(q=0;q<32;q++)
{
if((x%2)!=(y%2))
z++;
x>>=1;
y>>=1;
}
printf(“%d\n”,z);
}
return 0;
}
[em:02]

我觉得位运算最省事

include

int main()
{
int i,t;
scanf(“%d”,&t);
for(i=0;i<t;i++)
{
long int a,b,c;
int n=0;
scanf(“%ld %ld”,&a,&b);
c=a^b;
while(c!=0)
{
n+=c%2;
c/=2;
}
printf(“%d\n”,n);
}
return 0;
}

include

using namespace std;
int main()
{
int T;
cin >> T;
while(T–){
int x, y;
cin >> x >> y;
int cnt = 0, z = x ^ y;
while(z){
cnt += z % 2;
z = z >> 1;
}
cout << cnt << endl;
}
}