一道囧的模拟题，可以这样做，赞。
大意是求出从1~n的罗马数字的各个字母的统计。
I, II, III, IV, V, VI, VII, VIII, IX 1~9X, XX, XXX, XL, L, LX, LXX, LXXX, XC 10~90C, CC, CCC, CD, D, DC, DCC, DCCC, CM 100~900M, MM, MMM 1000~3000
然后用这些数字组合。
通过这题算是认识罗马数字了。
1 /* 2 ID: litstrong 3 PROG: preface 4 LANG: C++ 5 */ 6 7 #include < iostream > 8 #include < string > 9 #include < algorithm > 10 #include < string .h > 11 #include < vector > 12 #include < math.h > 13 #include < map > 14 #include < time.h > 15 #include < queue > 16 #include < set > 17 using namespace std; 18 19 /* 20 I 1 L 50 M 1000 21 V 5 C 100 22 X 10 D 500 23 I, II, III, IV, V, VI, VII, VIII, IX 1~9 24 X, XX, XXX, XL, L, LX, LXX, LXXX, XC 10~90 25 C, CC, CCC, CD, D, DC, DCC, DCCC, CM 100~900 26 M, MM, MMM 1000~3000 27 */ 28 char a[ 10 ][ 10 ] = { " I " , " II " , " III " , " IV " , " V " , " VI " , " VII " , " VIII " , " IX " }; 29 char b[ 10 ][ 10 ] = { " X " , " XX " , " XXX " , " XL " , " L " , " LX " , " LXX " , " LXXX " , " XC " }; 30 char c[ 10 ][ 10 ] = { " C " , " CC " , " CCC " , " CD " , " D " , " DC " , " DCC " , " DCCC " , " CM " }; 31 char d[ 10 ][ 10 ] = { " M " , " MM " , " MMM " }; 32 char mo[ 10 ] = " IVXLCDM " ; 33 34 int n; 35 map < char , int > ci; 36 37 void compute( char * ch) 38 { 39 int len = strlen(ch); 40 for ( int i = 0 ; i < len; i ++ ) 41 { 42 ci[ch[i]] ++ ; 43 } 44 } 45 46 void cnt( int val) 47 { 48 int t = 0 ; 49 while (val) 50 { 51 int tt = val % 10 ; 52 val /= 10 ; 53 t ++ ; 54 if (tt == 0 ) continue ; 55 if (t == 1 ) compute(a[tt - 1 ]); 56 else if (t == 2 ) compute(b[tt - 1 ]); 57 else if (t == 3 ) compute(c[tt - 1 ]); 58 else if (t == 4 ) compute(d[tt - 1 ]); 59 } 60 } 61 62 void go() 63 { 64 for ( int i = 1 ; i <= n; i ++ ) 65 cnt(i); 66 int len = strlen(mo); 67 for ( int i = 0 ; i < len; i ++ ) 68 if (ci[mo[i]] > 0 ) 69 printf( " %c %d\n " , mo[i], ci[mo[i]]); 70 } 71 72 void ready() 73 { 74 for ( int i = 0 ; i < strlen(mo); i ++ ) 75 { 76 ci[mo[i]] = 0 ; 77 } 78 } 79 80 int main() 81 { 82 freopen( " preface.in " , " r " , stdin); 83 freopen( " preface.out " , " w " , stdout); 84 85 scanf( " %d " , & n); 86 ready(); 87 go(); 88 }
转载于:https://www.cnblogs.com/litstrong/archive/2010/04/28/1723281.html
