C语言竞赛初级组第一、二场答案：https://www.cnblogs.com/xingkongyihao/p/10046918.html 

A: 逆序对

时间限制: 1 s      内存限制: 128 MB 

题目描述

1.整数序列中两个相邻的数，如果后面的数小于前面的数，则称这两个数值构成了一个逆序对。例如，整数序列10，4，16，8，21，18，9中包含了4个逆序对。从键盘上输入n个由空格分隔的整数，编程输出其中包含的逆序对的数量。

输入

第一行输入一个数字n (1≤n≤1000)

第二行输入n个由空格分隔的整数

输出

输出一个数字，表示逆序对的答案

样例输入

7

10 4 16 8 21 18 9

样例输出

4

果然就是比较相邻么
using System;

using System.Collections;

using System.Collections.Generic;

using System.IO;

using System.Linq;

using System.Runtime.InteropServices;

public class Program

{

    public static void Main()

    {

        int cunt = 0;

        int n = ri();

        List<int> a = new List<string>(Console.ReadLine().Split()).ConvertAll(i => int.Parse(i));

        for (int i = 1; i < n; i++)

        {

            if (a[i] < a[i - 1])

            {

                cunt++;

            }

        }

        Console.WriteLine(cunt);

        Console.ReadKey();

    }

    public static int ri() { return int.Parse(Console.ReadLine()); }

    static int[] rla(char sep = ' ') { return Array.ConvertAll(Console.ReadLine().Split(sep), e =>int.Parse(e)); }

}

B: sky的圈圈

时间限制: 1 s      内存限制: 128 MB     

题目描述

最近小S不高兴了，所以她就想画圈圈，有大有小的。而现在她想让你也画圈圈了^_^。

大小为3的圈圈是aaarticlea/png;base64,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" alt="" />，大小为4的圈圈是aaarticlea/png;base64,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" alt="" />，大小为5的圈圈是aaarticlea/png;base64,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" alt="" />，依次类推。

输入

输入一个数字n (3≤n≤100)

输出

输出你画的圈圈。

样例输入

3

样例输出

*#*

#*#

*#*

画图画图。。
using System;

using System.IO;

using System.Linq;

public class Program

{

    public static void Main()

    {

        var n = ri();

        char[,] Map = new char[200,200];

        Map[0, 0] = '*';

        Map[0, n - 1] = '*';

        Map[n - 1, 0] = '*';

        Map[n - 1, n - 1] = '*';

        for (int i = 1; i < n-1; i++)

        {

            Map[0, i] = '#';

            Map[n - 1, i] = '#';

        }

        for (int i = 1; i < n - 1; i++)

        {

            Map[i, 0] = '#';

            Map[i, n-1] = '#';

        }

        for (int i = 1; i < n - 1; i++)

        {

            for (int j = 1; j < n - 1; j++)

            {

                Map[i, j] = '*';

            }

        }

        for (int i = 0; i < n; i++)

        {

            for (int j = 0; j < n; j++)

            {

                Console.Write(Map[i,j]);

            }

            Console.WriteLine();

        }

        Console.ReadKey();

    }

    public static int ri() { return int.Parse(Console.ReadLine()); }

    static int[] rla(char sep = ' ') { return Array.ConvertAll(Console.ReadLine().Split(sep), e =>int.Parse(e)); }

}

C: 找零钱

时间限制: 1 s      内存限制: 128 MB    

题目描述

小明现在有x元，现在想买一件y (y≤x)元的物品，商店里有五种货币，100元、20元、10元、5元、1元无限张，服务员会以最小的数量找零钱。问小明用x元买了一件y元的物品后找了多少张零钱。

输入

输入x和y两个整数(1≤y≤x≤10000)

输出

输出找零钱的最小的数量

样例输入

101 66

样例输出

3

提示

用101元买了一件66元的物品，需要找35元，一张20元、一张10元和一张5元。

贪心，取模取模取模

using System;

using System.Collections;

using System.Collections.Generic;

using System.IO;

using System.Linq;

using System.Runtime.InteropServices;

public class Program

{

    public static void Main()

    {

        int cunt;

        int result = 0;

        List<int> a = new List<string>(Console.ReadLine().Split()).ConvertAll(i => int.Parse(i));

        cunt = a[0] - a[1];

        result += cunt / 100;

        cunt %= 100;

        result += cunt / 20;

        cunt %= 20;

        result += cunt / 10;

        cunt %= 10;

        result += cunt/5;

        cunt %= 5;

        result += cunt / 1;

        Console.WriteLine(result);

        Console.ReadKey();

    }

    public static int ri() { return int.Parse(Console.ReadLine()); }

    static int[] rla(char sep = ' ') { return Array.ConvertAll(Console.ReadLine().Split(sep), e =>int.Parse(e)); }

}

D: 编程语言

时间限制: 1 s      内存限制: 128 MB  

题目描述

现在的编程越来越多了，比如C、C++、Java、Python、C#、PHP等等。现在给定n种编程语言，每种语言还会给一个[1,n]之间的随机值，保证不重复。现在让你按随机值从小到大排序，然后输出对应的语言。

输入

第一行输入一个整数n (1≤n≤20)

接下来nn行，每行有一个字符串和一个随机值，字符串表示一种语言，长度不超过20.随机值范围为[1,n]

输出

输出nn行，按随机值从小到大输出对应的语言

样例输入

4

Java 3

C 1

Python 4

C++ 2

样例输出

C

C++

Java

Python

排序排序排序排序
using System;

using System.Collections;

using System.IO;

using System.Linq;

using System.Runtime.InteropServices;

public class Program

{

    public static void Main()

    {

        var n = ri();

        ArrayList al = new ArrayList();

        for (int i = 0; i < n; i++)

        {

            Language La = new Language();

            var x = Console.ReadLine().Split();

            La.xx = x[0];

            La.yy = int.Parse(x[1]);

            al.Add(La);

        }

        Languagex languagex = new Languagex();

        al.Sort(languagex);

        Language Laa = new Language();

        for (int i = 0; i < n; i++)

        {

            Laa = (Language) al[i];

            Console.WriteLine(Laa.xx);

        }

        Console.ReadKey();

    }

    public class Language

    {

        public string xx;

        public int yy;

    }

    public class Languagex:IComparer

    {

        public int Compare(object x, object y)

        {

            return ((Language) x).yy.CompareTo(((Language) y).yy);

        }

    }

    public static int ri() { return int.Parse(Console.ReadLine()); }

    static int[] rla(char sep = ' ') { return Array.ConvertAll(Console.ReadLine().Split(sep), e =>int.Parse(e)); }

}

E: 四边形面积

时间限制: 1 s      内存限制: 128 MB    

题目描述

有一个四边形，现在需要求它的面积

输入

输入四行，每行两个数整数x, y (1≤x,y≤1000)，四个点是按逆时针输入的。

输出

输出四边形的面积，保留3位小数点，

样例输入

0 0

10 0

1 1

0 11

样例输出

10.500

提示

C语言中保留3位小数用%.3lf   用法：printf("%.3lf",result)

四边形分凸凹四边形。

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" alt="" />

提示都告诉你了

#include<bits/stdc++.h>

using namespace std;

struct Point

{

    float x, y;

};

float LinearIntegration(const Point &p1, const Point &p2)

{

    return 0.5 * (p2.x - p1.x) * (p2.y + p1.y);

}

float ComputePolygonArea(const Point points[], int length)

{

    if (points == NULL || length <= 0) return 0.0;

    float area = 0.0;

    for (int i = 0; i < length - 1; ++ i)

    {

        area += LinearIntegration(points[i], points[i + 1]);

    }

    area += LinearIntegration(points[length - 1], points[0]);

    return area >= 0.0 ? area : -area;

}

int main()

{

    int n;

    Point a[4];

    for(int i=0; i<4;i++) cin>>a[i].x>>a[i].y;

    float ans = ComputePolygonArea(a,4);

    printf("%.3f\n",ans);

    return 0;

}

F: 进制转换

时间限制: 1 s      内存限制: 128 MB     

题目描述

给定一个区间[l, r]，从l至r之间的所有数依次转换成16进制然后连在一起，接着再转换成10进制，最后再对15取模。

输入

输入两个是l, r (1≤l≤r≤106)

输出

输出对15取模的结果。

样例输入

10 14

样例输出

0

提示

样例说明：

10、11、12、13、14的16进制分别是a、b、c、d、e。依次连在一起是abcde，转换成10进制是703710，对15取模为0。

一般的转化。(a*16^n+b*16^(n-1)..)%15，展开，a%16*16^n%15，16%15就可以不考虑了，于是变成a%15+b%15..。于是简化成a+b+c+d+e。注意取模

using System;

using System.Collections;

using System.Collections.Generic;

using System.IO;

using System.Linq;

using System.Reflection.Emit;

using System.Runtime.InteropServices;

public class Program

{

    public static void Main()

    {

        var str = "";

        List<int> a = new List<string>(Console.ReadLine().Split()).ConvertAll(i => int.Parse(i));

        var sum = 0;

        for (int i = a[0]; i <= a[1]; i++)

        {

            var pa = Convert.ToString(i, 16);

            //var pa = Convert.ToInt32(x, 16).ToString();

            var len = pa.Length;

            for (int j = 0; j < len; j++)

            {

                if (pa[j] == 'a')

                {

                    sum += (10);

                    sum %= 15;

                }

                else if (pa[j] == 'b')

                {

                    sum += (11);

                    sum %= 15;

                }

                else if (pa[j] == 'c')

                {

                    sum += (12);

                    sum %= 15;

                }

                else if (pa[j] == 'd')

                {

                    sum += (13);

                    sum %= 15;

                }

                else if (pa[j] == 'e')

                {

                    sum += (14);

                    sum %= 15;

                }

                else if (pa[j] == 'f')

                {

                    sum += (15);

                    sum %= 15;

                }

                else

                {

                    sum += int.Parse(pa[j].ToString());

                    sum %= 15;

                }

            }

        }

        Console.WriteLine(sum%15);

        Console.ReadKey();

    }

    public static int ri() { return int.Parse(Console.ReadLine()); }

    static int[] rla(char sep = ' ') { return Array.ConvertAll(Console.ReadLine().Split(sep), e =>int.Parse(e)); }

}

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