Questo è un esercizio del genere che stampa i n romani da 1 a 100.
Ciaocodice:#include<stdio.h> int main( void ) { int loop, div, mod; printf( " Roman\nNumeral\t\tDecimal\n" ); for ( loop = 1; loop <= 100; loop++ ) { div = loop / 10; mod = loop % 10; /* switch structure for tens digit */ switch ( div ) { case 0: break; case 1: printf( "X" ); break; case 2: printf( "XX" ); break; case 3: printf( "XXX" ); break; case 4: printf( "XL" ); break; case 5: printf( "L" ); break; case 6: printf( "LX" ); break; case 7: printf( "LXX" ); break; case 8: printf( "LXXX" ); break; case 9: printf( "XC" ); break; case 10: printf( "C" ); break; default: break; } /* switch structure for ones digit */ switch( mod ) { case 0: printf( "\t\t%4d\n", div * 10 ); break; case 1: printf( "I\t\t%4d\n", div * 10 + mod ); break; case 2: printf( "II\t\t%4d\n", div * 10 + mod ); break; case 3: printf( "III\t\t%4d\n", div * 10 + mod ); break; case 4: printf( "IV\t\t%4d\n", div * 10 + mod ); break; case 5: printf( "V\t\t%4d\n", div * 10 + mod ); break; case 6: printf( "VI\t\t%4d\n", div * 10 + mod ); break; case 7: printf( "VII\t\t%4d\n", div * 10 + mod ); break; case 8: printf( "VIII\t\t%4d\n", div * 10 + mod ); break; case 9: printf( "IX\t\t%4d\n", div * 10 + mod ); break; case 10: printf( "X\t\t%4d\n", div * 10 + mod ); break; default: break; } } system ("PAUSE"); return 0; }![]()

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