zoukankan      html  css  js  c++  java
  • E

    Description

    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 integerst1, 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 Input

    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
    #include<stdio.h>
    int a[5500];
    int min(int a,int b)
    {
        int c;
        b<a?c=b:c=a;
        return c;
    }
    void shuchu(int a[],int j,int n)
    {
        int i,t;
        for(i=0; i<j; i++)
        {
            for(t=0; t<n; t++)
            {
                if(a[t]==1)
                {
                    printf("%d ",t+1);
                    a[t]=0;
                    break;
                }
            }
            for(t=0; t<n; t++)
            {
                if(a[t]==2)
                {
                    printf("%d ",t+1);
                    a[t]=0;
                    break;
                }
            }
            for(t=0; t<n; t++)
            {
                if(a[t]==3)
                {
                    printf("%d
    ",t+1);
                    a[t]=0;
                    break;
                }
            }
        }
    }
    int main()
    {
        int i,j,t,n,a1,b1,c1;
        scanf("%d",&n);
        for(i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        a1=0;b1=0;c1=0;
        for(i=0; i<n; i++)
        {
            if(a[i]==1)a1++;
            else if(a[i]==2)b1++;
            else  c1++;
        }
        j=min(a1,min(b1,c1));
        if(j==0)printf("0
    ");
        else
            printf("%d
    ",j);
            shuchu(a,j,n);
        return 0;
    }
  • 相关阅读:
    drawcall优化
    Java基础:动态代理在RPC框架中应用
    ACdream ACfun
    图片压缩之处理小图片压缩策略
    ASP内置对象—Request、Response 、Server、Application 、ObjectContent(一)
    alter
    UltraISO制作U盘启动盘安装Win7/8/10系统攻略
    Libcurl的初步实现tfp上传下载功能
    微信企业号开发:获取AccessToken
    Ubuntu 安装OpenGL
  • 原文地址:https://www.cnblogs.com/dongq/p/4134014.html
Copyright © 2011-2022 走看看