金色十月線上編程比賽第二題：解密

金色十月線上編程比賽第二題：解密

金色十月線上編程比賽第二題：解密

• 發布公司：
• 有 效 期：
  • CSDN
  • 2014-10-20至2015-10-20
    • 難 度 等 級：
    • 答 題 時 長：
    • 編程語言要求：
      • 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"ans_tip">
        

        解題思路：利用這個函數就可以計算一個數的二進制有多少個1：

        int totalOne(int x){
            int count = 0;
            while(x){
                x = x & ( x - 1 );
                count++;
            }
            return count;
        }

        所以我們只需遍歷一邊即可；
        

        #include 
        using namespace std;
        int totalOne(int x){
            int count = 0;
            while(x){
                x = x & ( x - 1 );
                count++;
            }
            return count;
        }
        int main(){
            int n,a,step=1;    
            while (cin>>n){
                int max=0,num=0,min=0xffffff;
                for (int i=0;i>a;
                    num=totalOne(a);
                    if (numa)
                        max=a;    
                }
                cout<<"Case "<