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:
    Flink 读取 Kafka 数据 (极简版)
    自动化测试模型
    C语言字符串处理库函数大全(转)
    c语言笔记
    c语言的自动类型转换(转)
    itest(爱测试)开源接口测试&敏捷测试管理平台8.1.0发布
    itest(爱测试)开源接口测试&敏捷测试&极简项目管理 8.0.0 发布,测试重大升级
    工控机折腾小记
    linux
    华为交换机服务端策略路由配置
  • 原文地址:https://www.cnblogs.com/dongq/p/4140324.html
Copyright © 2011-2022 走看看