2012年北理复试上机题

#include <cstdio>
#include <algorithm>
using namespace std;

int main()
{
	int i,a[10];
	for(i=0;i<10;i++)
	{
		scanf("%d",&a[i]);

	}
	sort(a,a+10);
	for(i=0;i<10;i++)
	{
		printf("%d ",a[i]);

	}
	printf("\n");


}

#include <cstdio>
#include <iostream>
#include <string>
#include <vector>
#include <algorithm>
#define MaxSize 100

using namespace std;
typedef struct student
{
	int num;
	string name;
	string sex;
	int age;
}student;

bool cmp(student a,student b)
{
	return a.num<=b.num;
}
void Print(vector<student> stu)
{
	vector<student>::iterator it;
	for(it=stu.begin();it!=stu.end();it++)
	{
		if(it!=stu.begin())
			cout<<" , ";
		cout<<"("<<it->num<<","<<it->name<<","<<it->sex<<","<<it->age<<")";

	}
}



int main()
{
	
	vector<student> stu;
	student a;
	a.num=10;a.name="wes";a.sex="f";a.age=23;
	stu.push_back(a);
	a.num=20;a.name="ert";a.sex="f";a.age=45;
	stu.push_back(a);
	a.num=30;a.name="str";a.sex="t";a.age=89;
	stu.push_back(a);
	Print(stu);
	char order[100];
	cout<<"\n请输入指令"<<endl;
	cin>>order;
	int temp;
	while(order!="#")
	{
		if(order[0]=='I')
		{
			char name[30];
			char sex[30];
			sscanf(order,"I%d,%[^,],%[^,],%d",&a.num,name,sex,&a.age);
			a.name=name;
			a.sex=sex;
//			sscanf(order,"[^,]",b);
//			cout<<b<<"\n"<<endl;
			stu.push_back(a);
		}
		else if(order[0]=='D')
		{
			sscanf(order,"D%d",&temp);
			vector<student>::iterator it;
			for(it=stu.begin();it!=stu.end();)
			{
				if(it->num==temp)
					stu.erase(it);
				else
					it++;
			}
		}
		sort(stu.begin(),stu.end(),cmp);
		Print(stu);
		cout<<"\n请输入指令"<<endl;
		cin>>order;

	}
	
	return 0;
}

 这题遇到了很多问题，首先是sscanf（）,在使用这个时，无论是前面的字符串还是后面用来接受的变量，注意都是字符数组，不要用string，但可以接收后，直接赋值给string。

(char []类型给string：直接赋值，反之不可，用for（i=0;i<str.length();i++） ch[i]=str[i];   结尾加上'\0')

当结构体类型是  迭代器 it，，访问它内容的时候就不要*，，，直接it->first.

vi.begin(),,括号勿忘。

#include <cstdio>
#include <iostream>
#include <string>
#define MaxSize 100010;

using namespace std;

string str1;
string str2;

void pre_order(int af,int ae,int bf,int be)   //af:后续起点，ae：后续终点，bf：终须起点，be：中序终点
{
	if(af>ae || bf>be )
		return ;
	if(af==ae && bf==be)
	{
		printf("%c",str1[af]);
		return ;
	}
	printf("%c",str1[ae]);
	for(int i=bf;i<=be;i++)
	{
		if(str2[i]==str1[ae])
			break;
	}
	pre_order(af,af+(i-bf)-1,bf,i-1);
	pre_order(af+(i-bf),ae-1,i+1,be);

}

int main()
{
	cout<<"输入（按后序，中序）"<<endl;
	cin>>str1>>str2;
	pre_order(0,str1.length()-1,0,str2.length()-1);
	cout<<endl;

	return 0;


}

 

#include <cstdio>
#include <iostream>
#include <string>
#define MaxSize 100010;

using namespace std;

string str1;
string str2;

void pre_order(int af,int ae,int bf,int be)   //af:后续起点，ae：后续终点，bf：终须起点，be：中序终点
{
	if(af>ae || bf>be )
		return ;
	if(af==ae && bf==be)
	{
		printf("%c",str1[af]);
		return ;
	}
	printf("%c",str1[ae]);
	for(int i=bf;i<=be;i++)
	{
		if(str2[i]==str1[ae])
			break;
	}
	pre_order(af,af+(i-bf)-1,bf,i-1);
	pre_order(af+(i-bf),ae-1,i+1,be);

}

void post_order(int af,int ae,int bf,int be) //a:先序，b:中序
{
	if(af>ae || bf > be)
		return ;
	for(int i=bf;str2[i]!=str1[af];i++);
	post_order(af+1,af+i-bf,bf,i-1);
	post_order(af+i-bf+1,ae,i+1,be);

	printf("%c",str1[af]);

}

int main()
{
/*	cout<<"输入（按后序，中序）"<<endl;
	cin>>str1>>str2;
	pre_order(0,str1.length()-1,0,str2.length()-1);
*/
	cout<<"输入（按先序，中序）"<<endl;
	cin>>str1>>str2;
	post_order(0,str1.length()-1,0,str2.length()-1);
	cout<<endl;

	return 0;


}

 

#include <cstdio>
#include <iostream>
#include <string>


using namespace std;

typedef struct node
{
	char ch;
	struct node* l;
	struct node* r;
}TNode,*Tree;

string str1;
string str2;

void DFS(Tree T)
{
	if(T==NULL)
	{
		return;
		cout<<"NULL";
	}
	printf("%c",T->ch);
	DFS(T->l);
	DFS(T->r);
}

Tree create(int af,int ae,int bf,int be)
{
	if(af>ae || bf>be )
		return NULL;
	Tree T=(Tree)malloc(sizeof(TNode));
	T->l=NULL;
	T->r=NULL;
	T->ch=str1[ae];
	for(int i=bf;i<=be;i++)
	{
		if(str2[i]==str1[ae])
			break;
	}
	T->l=create(af,af+(i-bf)-1,bf,i-1);
	T->r=create(af+(i-bf),ae-1,i+1,be);
	return T;

}


int main()
{
	Tree T;
	cout<<"输入（按后序，中序）"<<endl;
	cin>>str1>>str2;
	T=create(0,str1.length()-1,0,str2.length()-1);
	DFS(T);


	return 0;


}