zoukankan      html  css  js  c++  java
  • C

    Description

    In this problem your goal is to sort an array consisting of n integers in at most n swaps. For the given array find the sequence of swaps that makes the array sorted in the non-descending order. Swaps are performed consecutively, one after another.

    Note that in this problem you do not have to minimize the number of swaps — your task is to find any sequence that is no longer than n.

    Input

    The first line of the input contains integer n (1 ≤ n ≤ 3000) — the number of array elements. The second line contains elements of array: a0, a1, ..., an - 1 ( - 109 ≤ ai ≤ 109), where ai is the i-th element of the array. The elements are numerated from 0 to n - 1from left to right. Some integers may appear in the array more than once.

    Output

    In the first line print k (0 ≤ k ≤ n) — the number of swaps. Next k lines must contain the descriptions of the k swaps, one per line. Each swap should be printed as a pair of integers ij (0 ≤ i, j ≤ n - 1), representing the swap of elements ai and aj. You can print indices in the pairs in any order. The swaps are performed in the order they appear in the output, from the first to the last. It is allowed to print i = jand swap the same pair of elements multiple times.

    If there are multiple answers, print any of them. It is guaranteed that at least one answer exists.

    Sample Input

    Input
    5
    5 2 5 1 4
    Output
    2
    0 3
    4 2
    Input
    6
    10 20 20 40 60 60
    Output
    0
    Input
    2
    101 100
    Output
    1
    0 1
    #include<stdio.h>
    int a[3010],b[3010],c[3010];
    void sortquickly(int a[],int lenth)
    {
        int mid;
        mid = a[0];
        int i,j;
        i=0;
        j=lenth-1;
        if(lenth>1)
        {
            while(i<j)
            {
                for(;j>i;j--)
                {
                    if(a[j]<mid)
                    {
                        a[i++]=a[j];
                        break;
                    }
                }
                for(;i<j;i++)
                {
                    if(a[i]>mid)
                    {
                        a[j--]=a[i];
                        break;
                    }
                }
            }
            a[i]=mid;
            sortquickly(a,i);
            sortquickly(a+i+1,lenth-i-1);
        }
    }
    
    void function(int a[],int b[],int n)
    {
        int k=0,i,j,t,m=0;
        for(i=0; i<n; i++)
        {
            if(a[i]!=b[i])
            {
                for(t=i+1; t<n; t++)
                {
                    if(a[i]==b[t])
                    {
                        c[m++]=i;
                        c[m++]=t;
                        j=b[i];
                        b[i]=b[t];
                        b[t]=j;
                        break;
                    }
                }
            }
        }
        if(m==0)printf("%d",m/2);
        else
        {
            printf("%d
    ",m/2);
            for(i=0; i<m;)
                printf("%d %d
    ",c[i++],c[i++]);
        }
    }
    int main()
    {
        int n,i,t;
        scanf("%d",&n);
        for(i=0; i<n; i++)
            scanf("%d",&a[i]);
        for(i=0; i<n; i++)
            b[i]=a[i];
        sortquickly(a,n);
        function(a,b,n);
        return 0;
    }
  • 相关阅读:
    对象属性编辑器中实现像Size,Rectangle等可以展开的属性
    远程办公产品风口会不会把SOHO自由职业吹起来
    项目加
    推荐几款免费又好用的项目管理工具
    Sprint Retrospective
    敏捷管理的大概背景和Scrum的特性
    推荐几款最好用的项目管理系统,即好用又免费
    项目管理的需求变更问题
    敏捷管理有一个原则就是:拥抱变化
    推荐5款体验最好的项目管理工具
  • 原文地址:https://www.cnblogs.com/dongq/p/4140324.html
Copyright © 2011-2022 走看看