USACOTrainning.Preface Numbering
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 1~9
X, XX, XXX, XL, L, LX, LXX, LXXX, XC 10~90
C, CC, CCC, CD, D, DC, DCC, DCCC, CM 100~900
M, 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
}