2012Hulu校园招聘笔试题

一、填空

侧重逻辑思维，没有语言、具体技术考察，大部分属于组合数学、算法。比较基本的知识点有二元树节点树、最小生成树、Hash函数常用方法等。

二、编程题

1、正整数剖分

2、AOE关键路径

3、二元树前序、中序求后序

4、大整数加

 

//正整数剖分
#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;
}

//二元树前序、中序打印后序
#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;
}

//大整数运算
#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;
}