Codeforces Round #279 (Div. 2) A

最近水水的开始写CF,感觉自己什么都不会,现在连水题都切不动了。。。。

写一下我做的两道水题。。

A. Team Olympiad
time limit per test
1 second
memory limit per test
256 megabytes
input
standard input
output
standard output

The School №0 of the capital of Berland has n children studying in it. All the children in this school are gifted: some of them are good at programming, some are good at maths, others are good at PE (Physical Education). Hence, for each child we know value ti:

  • ti = 1, if the i-th child is good at programming,
  • ti = 2, if the i-th child is good at maths,
  • ti = 3, if the i-th child is good at PE

Each child happens to be good at exactly one of these three subjects.

The Team Scientific Decathlon Olympias requires teams of three students. The school teachers decided that the teams will be composed of three children that are good at different subjects. That is, each team must have one mathematician, one programmer and one sportsman. Of course, each child can be a member of no more than one team.

What is the maximum number of teams that the school will be able to present at the Olympiad? How should the teams be formed for that?

Input

The first line contains integer n (1 ≤ n ≤ 5000) — the number of children in the school. The second line contains n integers t1, t2, ..., tn(1 ≤ ti ≤ 3), where ti describes the skill of the i-th child.

Output

In the first line output integer w — the largest possible number of teams.

Then print w lines, containing three numbers in each line. Each triple represents the indexes of the children forming the team. You can print both the teams, and the numbers in the triplets in any order. The children are numbered from 1 to n in the order of their appearance in the input. Each child must participate in no more than one team. If there are several solutions, print any of them.

If no teams can be compiled, print the only line with value w equal to 0.

Sample test(s)
input
7
1 3 1 3 2 1 2
output
2
3 5 2
6 7 4
input
4
2 1 1 2
output
0

题目意思就是选出最多的含有1,2,3的种数,然后输出它们各自的下标。题目很简单,用一个二维数组M[4][N]和d[4]来实现,M的行用来读入1,2,3,M的列用来记录出现的次数,每次出现一次的d[k]++。然后用用ans找到1,2,3出现的次数最小的值,然后输出。
 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <cmath>
 5 
 6 using namespace std;
 7 
 8 
 9 
10 #define N 5010
11 
12 int m[4][N], d[4];
13 int n;
14 
15 int main() {
16 
17   scanf("%d", &n);
18   for (int i = 1; i <= n; i++) {
19     int k;
20     scanf("%d", &ff);
21     man[k][d[k]] = i;
22     d[k]++;
23 
24 
25   }
26   int ans = min(d[1], min(d[2], d[3]));
27   printf("%d\n", ans);
28   for (int i = 0; i < ans; i++)
29     printf("%d %d %d\n", m[1][i], m[2][i], m[3][i]);
30   return 0;
31 }

 

posted @ 2014-11-25 12:40  lmlyzxiao  阅读(158)  评论(0编辑  收藏  举报