UVa 537 - Artificial Intelligence?

Problem link

Keyword: ad hoc

Algorithm:

Since the equal sign will appear only at the relevant digits, we can scan the word one by one and see if there's an equal sign in it.
If yes, we can extract what we want and print the answer.
Notice the modifiers like "m" or "M."

Code:

#include<cstdio>
#include<cstring>

int main()
{
    int t;
    scanf( "%d\n", &t );

    for( int i = 1; i <= t; ++i )
    {
        bool current = false, voltage = false, power = false;
        char str[ 100000 ] = { 0 }, *pos = str, tmp[ 100000 ] = { 0 }, scale;
        double A, V, W;
        gets( str );
        while( *pos )
        {
            sscanf( pos, "%s", tmp );
            pos += strlen( tmp ) + 1;
            if( strchr( tmp, '=' ) )
                if( *tmp == 'U' )
                {
                    voltage = true;
                    sscanf( tmp+2, "%lf%c", &V, &scale );
                    if( scale == 'm' )
                        V *= 1e-3;
                    else if( scale == 'M' )
                        V *= 1e6;
                    else if( scale == 'k' )
                        V *= 1e3;
                }
                else if( *tmp == 'I' )
                {
                    current = true;
                    sscanf( tmp+2, "%lf%c", &A, &scale );
                    if( scale == 'm' )
                        A *= 1e-3;
                    else if( scale == 'M' )
                        A *= 1e6;
                    else if( scale == 'k' )
                        A *= 1e3;
                }
                else
                {
                    power = true;
                    sscanf( tmp+2, "%lf%c", &W, &scale );
                    if( scale == 'm' )
                        W *= 1e-3;
                    else if( scale == 'M' )
                        W *= 1e6;
                    else if( scale == 'k' )
                        W *= 1e3;
                }
        }
        printf( "Problem #%d\n", i );
        if( !current )
            printf( "I=%.2fA\n\n", W / V );
        else if( !voltage )
            printf( "U=%.2fV\n\n", W / A );
        else
            printf( "P=%.2fW\n\n", A * V );
    }

    return 0;
}