Question

神秘的表达式 求VC++6.0代码

Answer 1

AC 了.

#include

using namespace std;

int main()

{

int Z,U,T;

int testCase;

cin>>testCase;

while( testCase-- )

{

cin>>Z>>U>>T;

if( Z + U == T ) putchar('+');

else if( Z-U == T ) putchar('-');

else if( (long long) Z * U == (long long ) T ) putchar('*');

else if( Z % U == 0 && Z/U == T ) putchar('\');

putchar('n');

}

return 0;

}