The beauty of programming-question about Chinese chess masters

The question requires that all valid locations of "" (a) and "" (B) be output, and only one byte storage variable can be used;

Solution 1:

 1 #include<iostream> 2 using namespace std; 3  4 int main() 5 { 6     int i=81; 7     while(i--) 8     { 9         if(i/9%3==i%9%3)10             continue;11         else12             cout<<i/9+1<<‘\t‘<<i%9+1<<endl;13     }14     return 0;15 }

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There are nine methods for a and B, So I = 81, for an integer I = I/9*9 + I % 9; I % 9 with I/9 change round, it is equivalent to a nested loop; an I is used to represent the changes of two values;

Of course, you can also use this method to implement multiple nesting;

Solution 2:

 1 #include<iostream> 2 using namespace std; 3  4 struct myBite 5 { 6     unsigned char a:4; 7     unsigned char b:4; 8 }; 9 10 int main(void)11 {12     myBite i;13     for(i.a=1;i.a<=9;++i.a)14         for(i.b=1;i.b<=9;++i.b)15             if(i.a%3!=i.b%3)16                 cout<<(int)i.a<<‘\t‘<<(int)i.b<<endl;17     return 0;18 }

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Refer to related bit operations

 

The beauty of programming-question about Chinese chess masters