HULU2012校招筆試體會

 一、填空

側重邏輯思維，沒有語言、具體技術考察，大部分屬於組合數學、算法。比較基本的知識點有二元樹節點樹、最小生成樹、Hash函數常用方法等。

二、編程題

1、正整數剖分

2、AOE關鍵路徑

3、二元樹前序、中序求後序

4、大整數加

參考代碼

//正整數剖分
#include <stdio.h>

int func(int n, int k, int max)
{
	int min = (int)((n+k-1)/k);
	if(k==1)
		return 1;
	int count = 0;
	for(int i=min;i<max;i++){
		count += func(n-i, k-1, max-i);
	}
	return count;
}

int main()
{
	int ans;
	int i = 10;
	//ans = func(10, 3, 10);
	ans = func(4, 3, 4);
	printf("%d\n", ans);

	return 0;
}

 

//AOE
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void AOE(int adj[][9], int n)
{
	int *e = (int*)malloc(sizeof(int)*n);
	int *l = (int*)malloc(sizeof(int)*n);
	e[0]=0;
	for(int i=1;i<n;i++){
		int max = 0;
		for(int j=0;j<i;j++){
			if(adj[j][i]!=0 && adj[j][i]+e[j]>max){
				max = adj[j][i] + e[j];
			}
		}
		e[i]=max;
	}
	l[n-1]=e[n-1];
	for(int i=n-2;i>=0;i--){
		int min = 999999;
		for(int j=n-1;j>i;j--){
			if(adj[i][j]!=0 && l[j]-adj[i][j]<min){
				min = l[j]-adj[i][j];
			}
		}
		l[i]=min;
	}
	for(int i=0;i<n;i++){
		if(e[i]==l[i])
			printf("%d\t", i+1);
	}
	printf("\n");
	free(e);
	free(l);
}
int main()
{
	const int size = 9;
	int adj[size][size];
	memset(adj, 0, sizeof(int)*size*size);
	adj[0][1] = 6, adj[0][2]=4, adj[0][3]=5;
	adj[1][4] = 1, adj[2][4]=1, adj[3][5]=2;
	adj[4][6] = 6, adj[4][7]=5, adj[5][7]=4;
	adj[6][8] = 2, adj[7][8]=4;

	AOE(adj, size);

	return 0;
}

 

//二元樹前序、中序打印後序
#include <stdio.h>
#include <cstring>
#include <stack>
using namespace std;
void dumpPost(const char* pre, const char* mid)
{
	int n = strlen(pre);
	if(n==1){
		printf("%c\t", pre[0]);
		return;
	}
	int i;
	for(i=0;i<n;i++){
		if(mid[i]==pre[0])
			break;
	}
	char lpre[i], lmid[n-i-1];
	char rpre[i], rmid[n-i-1];
	memcpy(lpre, pre+1, sizeof(char)*i);
	memcpy(lmid, mid, sizeof(char)*i);
	memcpy(rpre, pre+i+1, sizeof(char)*(n-i-1));
	memcpy(rmid, mid+i+1, sizeof(char)*(n-i-1));
	lpre[i] = lmid[i] = '\0';
	rpre[n-i-1] = rmid[n-i-1] = '\0';
	dumpPost(lpre, lmid);
	dumpPost(rpre, rmid);
	printf("%c\t", pre[0]);
}
int main()
{
	const char* preOrder = "ABDEC";
	const char* midOrder = "DBEAC";
	const char* postOrder = "DEBCA";

	dumpPost(preOrder, midOrder);
	printf("\n");

	return 0;
}

 

//大整數運算
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

void strrev(char* s)
{
	int i=-1;
	while(s[++i]!='\0');
	for(int j=0;j<i/2;j++){
		char tmp = s[j];
		s[j] = s[i-j-1];
		s[i-j-1]=tmp;
	}
}
void Add(const char*str1, const char* str2, char* ans)
{
	int l1, l2, l;
	l1 = strlen(str1);
	l2 = strlen(str2);
	l = l1>l2 ? l1 : l2;
	char* s1 = (char*)malloc(sizeof(char)*(l1+1));
	char* s2 = (char*)malloc(sizeof(char)*(l2+1));
	memcpy(s1,str1,(l1+1)*sizeof(char));
	memcpy(s2,str2,(l2+1)*sizeof(char));
	strrev(s1);
	strrev(s2);
	int i;
	int sum, carry;
	i=sum=carry=0;
	while(i<l){
		char a = i<l1?s1[i]:'0';
		char b = i<l2?s2[i]:'0';
		sum = a-'0'+b-'0' + carry;
		ans[i] = sum % 10 + '0';
		carry = sum / 10;
		i++;
	}
	if(carry!=0)
		ans[i++]=carry+'0';
	ans[i]='\0';
	strrev(ans);
	free(s1);
	free(s2);
}
void Mul(const char*str1, const char* str2, char* ans)
{
	int l1, l2, l;
	l1 = strlen(str1);
	l2 = strlen(str2);
	l = l1 + l2;
	ans[0]='\0';
	char* s1 = (char*)malloc(sizeof(char)*(l1+1));
	char* s2 = (char*)malloc(sizeof(char)*(l2+1));
	memcpy(s1,str1,(l1+1)*sizeof(char));
	memcpy(s2,str2,(l2+1)*sizeof(char));
	strrev(s1);
	strrev(s2);
	char* tmp = (char*)malloc(sizeof(char)*(l1+2));
	int s, carry;
	s = carry = 0;
	for(int i=0;i<l2;i++){
		int j;
		for(int j=0;j<i;j++)
			tmp[j]='0';
		for(j=0;j<l1;j++){
			s = (s1[j]-'0')*(s2[i]-'0')+carry;
			tmp[i+j]=s%10+'0';
			carry=s/10;
		}
		if(carry!=0)
			tmp[i+j++]=carry+'0';
		tmp[i+j]='\0';
		strrev(ans);
		strrev(tmp);
		Add(ans,tmp, ans);
		strrev(ans);
	}
	strrev(ans);
}
int main()
{
	const char a[] = "12345";
	const char b[] = "123";
	char c[1024];

	Add(a,b,c);
	printf("a+b=%s\n", c);
	Mul(a,b,c);
	printf("a*b=%s\n", c);

	return 0;
}