程序设计初步ppt和训练题

末蓝、 2023-07-03 08:26 10阅读 0赞

本次的内容是依学校校队的安排，给学弟学妹们讲解程序设计初步的ppt，因为偏基础，只是讲的话效果不是很好，因此在VJ上挂了一套题，感兴趣的同学可以到VJ上自己做着玩。链接如下： [https://vjudge.net/contest/355440][https_vjudge.net_contest_355440]

PPT的下载地址：[https://download.csdn.net/download/zhiyeegao/12136528][https_download.csdn.net_download_zhiyeegao_12136528]

PPT如下：

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXllZWdhbw_size_16_color_FFFFFF_t_70][]

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![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXllZWdhbw_size_16_color_FFFFFF_t_70 2][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXllZWdhbw_size_16_color_FFFFFF_t_70 3][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXllZWdhbw_size_16_color_FFFFFF_t_70 4][]

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![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXllZWdhbw_size_16_color_FFFFFF_t_70 7][]

![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXllZWdhbw_size_16_color_FFFFFF_t_70 8][]

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![watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3poaXllZWdhbw_size_16_color_FFFFFF_t_70 11][]

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题解如下：（考虑到基础性问题，题解中最难就是循环，十分适合新手做）

A．	水仙花数
    
    #include<iostream>
    #include<stdio.h>
    #include<math.h>
    #include<stdlib.h>
    #include<string.h>
    using namespace std;
    
    int main()
    {
        int n,m;
        while(~scanf("%d%d",&n,&m))
        {
            int t;
            if(n>m)
            {
                t=n;
                n=m;
                m=t;
            }
            int flag=0;
            for(int i=n;i<=m;i++)
            {
                int a=i/100;
                int b=i%10;
                int c=i/10%10;
                if(i==a*a*a+b*b*b+c*c*c)
                {
                    if(flag==0)
                    {
                        printf("%d",i);
                        flag=1;
                    }
                    else
                    {
                        printf(" %d",i);
                    }
                }
            }
            if(flag==0)
            printf("no");
            printf("\n");
        }
        return 0;
    }
    
    B．	A+B
    
    #include<iostream>
    using namespace std;
    int main()
    {
    	int n;
    	cin>>n;
    	while(n--){
    		long long s,a,b;
    		cin>>a>>b;
    		s=a+b;
    		s=s%100;
    		cout<<s<<endl;
    	}
     }
    
    C．	偶数拆分
    
    #include<stdio.h>
    #include<math.h>
    int is (int x)
    {
    	int i;
    	if(x==2) return 1;
    	for(i=2;i<=sqrt(x);i++)
    		if(x%i==0)
    			return 0;
    	return 1;
    }
    int main()
    {
    	int i,n,m;
    	while(1)
    	{
    		n=0;
    		scanf("%d",&m);
    		if(m==0) break;
    		for(i=2;i<m-i;i++)
    			if(is(i)&&is(m-i))
    				n++;
    		printf("%d\n",n);
    	}
    	return 0;
    }
    D．	牛生牛
    //类似斐波拉契：old（第n-1月的牛数）+new（新生牛数，易知新生牛数量等于第n-3月牛数）
    #include<iostream>
    using namespace std;
    int fibonacci1(int n)
    {
        switch(n)
        {
            case 1:return 1;
            case 2:return 2;
            case 3:return 3;
            default:return fibonacci1(n-3)+fibonacci1(n-1);
        }
    }
    int main()
    {
        int n;
        while(cin>>n&&n)
        {
            cout<<fibonacci1(n)<<endl;
        }
        return 0;
    }
    E．	一根绳上的蚂蚁
    
    #include<cstdio>
    #include<algorithm>
    using namespace std;
    int main()
    {
        int T,ans,sum,a,N,M;
        scanf("%d",&T);
        while(T--)
        {
            scanf("%d%d",&N,&M);
            ans=sum=0;
            while(M--)
            {
                scanf("%d",&a);
                ans=max(ans,max(a,N-a));
                sum=max(sum,min(a,N-a));
            }
            printf("%d %d\n",sum,ans);
        }
        return 0;
    }
    
    F．	魅力序列
    
    #include<iostream> 
    #include<string> 
    #include<algorithm> 
    using namespace std; 
    int main(void) 
    { 
        long long n,m,e; 
        long long ans; 
        while (~scanf("%lld%lld",&n,&m)) 
        { 
            e=n-n%(2*m); 
            ans=(e)*(-m)/2;//所有完整段求和
            if(e+m<n)//分类讨论
            {
            	ans=ans+(2*e+1+m)*m/2;
            	ans=ans-(e+m+1+n)*(n-e-m)/2;
            }
            else if(e+m>n)
            {
            	ans=ans+(e+1+n)*(n-e)/2;
            }
            else
            {
            	ans=ans+(2*e+1+m)*m/2;
            }
        	printf("%lld\n",ans);//这题要用long long来储存答案
        } 
        return 0; 
    }   
    
    可以发现从1开始前2*m项的和的绝对值是一个常数即m，e=n-n%(2*m)那么只需处理后面e+1~n的数字即可。因此可以有如下规律
    
    设剩下的区间为e+1~n（也可能是e+m）。
    
    1、若e+m<=n——即数轴分布情况为e+1~e+m~n。只需将之前的所有完整段的和加上Sum[e+1,e+m]即可，且每一项都是正数。
    
    2、若e+m>n——即数轴分布情况为e+1~n~e+m。那么要计算S1[e+1,n]与S2[n+1,e+m]，ans=ans+S1-S2。

[https_vjudge.net_contest_355440]: https://vjudge.net/contest/355440
[https_download.csdn.net_download_zhiyeegao_12136528]: https://download.csdn.net/download/zhiyeegao/12136528
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