AtCoder Beginner Contest 286 题目讲解
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初中生蒟蒻
今天讲解一下AtCoder Beginner Contest 286的A题——Range Swap
===========================================================================================
原题
Problem Statement
You are given a sequence
A
=
(
A
1
,
A
2
,
…
,
A
N
)
A = (A_{1}, A_{2}, … , A_{N})
A
=
(
A
1
,
A
2
,
…
,
A
N
)
of length
N
N
N
and positive integers
P
,
Q
,
R
P,Q,R
P
,
Q
,
R
, and
S
S
S
.
Here,
P
P
P
,
Q
Q
Q
,
R
R
R
, and
S
S
S
satisfy
1
≤
P
≤
Q
<
R
≤
S
≤
N
1\leq P\leq Q < R\leq S\leq N
1
≤
P
≤
Q
<
R
≤
S
≤
N
and
Q
−
P
=
S
−
R
.
Q−P=S−R.
Q
−
P
=
S
−
R
.
Let
B
=
(
B
1
,
B
2
,
…
,
B
N
)
B=(B_{1},B_{2}, …,B_{N})
B
=
(
B
1
,
B
2
,
…
,
B
N
)
be the sequence obtained by swapping the
P
P
P
-th through
Q
Q
Q
-th terms and the
R
R
R
-th through
S
S
S
-th terms of
A
A
A
.
Print the sequence
B
B
B
.
Constraints
-
1≤
N
≤
100
1\leq N\leq 100
1
≤
N
≤
100
-
1≤
A
i
≤
100
1\leq A_{i}\leq 100
1
≤
A
i
≤
100
-
1≤
P
≤
Q
<
R
≤
S
≤
N
1\leq P\leq Q<R\leq S\leq N
1
≤
P
≤
Q
<
R
≤
S
≤
N
-
Q−
P
=
S
−
R
Q−P=S−R
Q
−
P
=
S
−
R
- All values in the input are integers.
Input
The input is given from Standard Input in the following format:
N P Q R S
A1 A2 … AN
Output
Print
B
1
,
B
2
,
…
,
B
N
B_{1},B_{2},…,B_{N}
B
1
,
B
2
,
…
,
B
N
, with spaces in between.
Sample Input 1
8 1 3 5 7
1 2 3 4 5 6 7 8
Sample Output 1
5 6 7 4 1 2 3 8
Swapping the
1
1
1
-st through
3
3
3
-rd terms
(
1
,
2
,
3
)
(1,2,3)
(
1
,
2
,
3
)
and the
5
5
5
-th through
7
7
7
-th terms
(
5
,
6
,
7
)
(5,6,7)
(
5
,
6
,
7
)
of the sequence
A
=
(
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
)
A=(1,2,3,4,5,6,7,8)
A
=
(
1
,
2
,
3
,
4
,
5
,
6
,
7
,
8
)
results in
B
=
(
5
,
6
,
7
,
4
,
1
,
2
,
3
,
8
)
B=(5,6,7,4,1,2,3,8)
B
=
(
5
,
6
,
7
,
4
,
1
,
2
,
3
,
8
)
, which should be printed with spaces in between.
Sample Input 2
5 2 3 4 5
2 2 1 1 1
Sample Output 2
2 1 1 2 1
The same integer may occur multiple times in the sequence.
Sample Input 3
2 1 1 2 2
50 100
Sample Output 3
100 50
Sample Input 4
10 2 4 7 9
22 75 26 45 72 81 47 29 97 2
Sample Output 4
22 47 29 97 72 81 75 26 45 2
思路
因为就只要更改2个区间，所以整个序列分成5个段，依次输出即可。（详情请看代码）
代码
#include <iostream>
using namespace std;
const int N = 1e2 + 10;
int n, p, q, r, s;
int a[N];
int main()
{
cin >> n >> p >> q >> r >> s;
for (int i = 1; i <= n; i ++)
cin >> a[i];
for (int i = 1; i < p; i ++)//区间1之前的位置
cout << a[i] << " ";
for (int i = r; i <= s; i ++) //区间1的位置输出区间2的元素的值
cout <<a[i] << " ";
for (int i = q + 1; i < r; i ++)//区间1与2之间的值
cout << a[i] << " ";
for (int i = p; i <= q; i ++)//区间2的位置输出区间1的元素的值
cout << a[i] << " ";
for (int i = s + 1; i <= n; i ++) //区间2之后的位置的值
cout << a[i] << " ";
return 0;
}
今天就到这里了！
大家有什么问题尽管提，我都会尽力回答的！最后，除夕夜祝大家新年快乐！
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