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

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

    return ;
}

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

void AOE(int adj[][], int n)
{
    int *e = (int*)malloc(sizeof(int)*n);
    int *l = (int*)malloc(sizeof(int)*n);
    e[]=;
    for(int i=;i<n;i++){
        int max = ;
        for(int j=;j<i;j++){
            if(adj[j][i]!= && adj[j][i]+e[j]>max){
                max = adj[j][i] + e[j];
            }
        }
        e[i]=max;
    }
    l[n-]=e[n-];
    for(int i=n-;i>=;i--){
        int min = ;
        for(int j=n-;j>i;j--){
            if(adj[i][j]!= && l[j]-adj[i][j]<min){
                min = l[j]-adj[i][j];
            }
        }
        l[i]=min;
    }
    for(int i=;i<n;i++){
        if(e[i]==l[i])
            printf("%d\t", i+);
    }
    printf("\n");
    free(e);
    free(l);
}
int main()
{
    const int size = ;
    int adj[size][size];
    memset(adj, , sizeof(int)*size*size);
    adj[][] = , adj[][]=, adj[][]=;
    adj[][] = , adj[][]=, adj[][]=;
    adj[][] = , adj[][]=, adj[][]=;
    adj[][] = , adj[][]=;

    AOE(adj, size);

    return ;
}

//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==){
        printf("%c\t", pre[]);
        return;
    }
    int i;
    for(i=;i<n;i++){
        if(mid[i]==pre[])
            break;
    }
    char lpre[i], lmid[n-i-];
    char rpre[i], rmid[n-i-];
    memcpy(lpre, pre+, sizeof(char)*i);
    memcpy(lmid, mid, sizeof(char)*i);
    memcpy(rpre, pre+i+, sizeof(char)*(n-i-));
    memcpy(rmid, mid+i+, sizeof(char)*(n-i-));
    lpre[i] = lmid[i] = '\0';
    rpre[n-i-] = rmid[n-i-] = '\0';
    dumpPost(lpre, lmid);
    dumpPost(rpre, rmid);
    printf("%c\t", pre[]);
}
int main()
{
    const char* preOrder = "ABDEC";
    const char* midOrder = "DBEAC";
    const char* postOrder = "DEBCA";

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

    return ;
}

//大整数运算
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

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

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

    return ;
}