题目
题目描述
Shuffling is
a procedure used
to randomize
a deck
of playing cards.
Because standard shuffling techniques are seen
as weak,
and in order
to
avoid
"inside jobs" where employees collaborate
with gamblers
by performing inadequate shuffles, many casinos employ automatic
shuffling machines
. Your task is
to simulate
a shuffling machine.
The machine shuffles
a deck
of 54 cards according
to a given
random
order
and repeats
for a given
number of times. It is assumed that
the
initial status
of a card deck is
in the following order:
S1, S2, ..., S13,
H1, H2, ..., H13,
C1, C2, ..., C13,
D1, D2, ..., D13,
J1, J2
where
"S" stands
for "Spade",
"H" for
"Heart",
"C" for "Club",
"D" for
"Diamond",
and "J" for "Joker". A given
order is
a permutation
of distinct integers
in [
1,
54]. If
the number
at the i-th position is j,
it means
to move
the card
from position i
to
position j. For example, suppose we only have
5 cards: S3, H5, C1, D13
and J2. Given
a shuffling order {
4,
2,
5,
3,
1},
the result will be:
J2, H5, D13, S3, C1. If we are
to repeat the shuffling again,
the
result will be: C1, H5, S3, J2, D13.
输入描述:
Each input
file contains one test
case. For
each case,
the first line contains a positive
integer K (<=
20) which is
the number of repeat times. Then
the next
line contains the given order. All
the numbers
in a line are separated
by a space.
输出描述:
For
each test
case, print
the shuffling results
in one line. All
the cards are separated
by a space,
and there must be no extra
space at the end of the line.
输入例子:
2
36 52 37 38 3 39 40 53 54 41 11 12 13 42 43 44 2 4 23 24 25 26 27 6 7 8 48 49 50 51 9 10 14 15 16 5 17 18 19 1 20 21 22 28 29 30 31 32 33 34 35 45 46 47
输出例子:
S7 C11 C10 C12 S1 H7 H8 H9 D8 D9 S11 S12 S13 D10 D11 D12 S3 S4 S6 S10 H1 H2 C13 D2 D3 D4 H6 H3 D13 J1 J2 C1 C2 C3 C4 D1 S5 H5 H11 H12 C6 C7 C8 C9 S2 S8 S9 H10 D5 D6 D7 H4 H13 C5
解题思路
1.用a[i]来保存原来的序列,其中i和a[i]都是1到54,给出一个orderb[i],题目的意思就是把1到54按照b[i]的序列来重置,我这里就用一个c[i],来保存a[i],因为要循环k次,然后根据b[i]来重置a[i] k次即可。
代码
#include<iostream>
#include<vector>
#include <map>
using namespace std;
int main()
{
vector<int> a(
55),b(
55),c(
55);
int k;
cin >> k;
for (
int i =
1; i <
55; i++)
{
a[i] = i;
cin >> b[i];
}
while (k--)
{
for (
int i =
1; i <
55; i++)
{
c[i] = a[i];
}
for (
int i =
1; i <
55; i++)
{
a[b[i]] = c[i];
}
}
for (
int i =
1; i <
55; i++)
{
int k1, k2;
k1 = a[i] /
13; k2 = a[i] %
13;
if (k2 ==
0)
{
k1--;
k2 =
13;
}
if (i==
1)
{
switch (k1)
{
case 0:
cout <<
"S" << k2;
break;
case 1:
cout <<
"H" << k2;
break;
case 2:
cout <<
"C" << k2;
break;
case 3:
cout <<
"D" << k2;
break;
case 4:
cout <<
"J" << k2;
}
}
else
{
cout <<
" ";
switch (k1)
{
case 0:
cout <<
"S" << k2;
break;
case 1:
cout <<
"H" << k2;
break;
case 2:
cout <<
"C" << k2;
break;
case 3:
cout <<
"D" << k2;
break;
case 4:
cout <<
"J" << k2;
}
}
}
cout << endl;
return 0;
}
转载请注明原文地址: https://ju.6miu.com/read-1284567.html