2020q3 Homework4 (quiz4)

contributed by < haoyu0970624763 >

測驗 1 - Hamming Distance

int hammingDistance(int x, int y){
    return __builtin_popcount(x ^ y);
}

程式原理

Hamming Distance 是要計算兩串二進位字串不同的地方有幾個，也就是計算 XOR 之後 有幾個 1

不使用 gcc extension 的 C99 實作程式碼

int hammingDistance(int x, int y) {
    x=x^y;
    unsigned count=0;
    while(x){
        count++;
        x=x&(x-1);
    }
    return count;
}

參考資訊

測驗二-Kth Ancestor of a Tree Node

#include <stdlib.h>

typedef struct {

    int **parent; 
    int max_level;
    
} TreeAncestor;

TreeAncestor *treeAncestorCreate(int n, int *parent, int parentSize)
{
    TreeAncestor *obj = malloc(sizeof(TreeAncestor));
    obj->parent = malloc(n * sizeof(int *));
    
    int max_level = 32 - __builtin_clz(n) + 1;
    for (int i = 0; i < n; i++) {
        obj->parent[i] = malloc(max_level * sizeof(int));
        for (int j = 0; j < max_level; j++)
            obj->parent[i][j] = -1;
    }
    for (int i = 0; i < parentSize; i++)
        obj->parent[i][0] = parent[i];

    for (int j = 1;; j++) {
        int quit = 1;
        for (int i = 0; i < parentSize; i++) {
            obj->parent[i][j] = obj->parent[i][j-1] == -1 ? -1:obj->parent[obj->parent[i][j - 1]];
            if (obj->parent[i][j] != -1) quit = 0;
        }
        if (quit) break;
    }
    obj->max_level = max_level - 1;
    return obj;
}

int treeAncestorGetKthAncestor(TreeAncestor *obj, int node, int k)
{
    int max_level = obj->max_level;
    for (int i = 0; i < max_level && node != -1; ++i)
        if (k & (1 << i))
            node = obj->parent[node][i];
    return node;
}

void treeAncestorFree(TreeAncestor *obj)
{
    for (int i = 0; i < obj->n; i++)
        free(obj->parent[i]);
    free(obj->parent);
    free(obj);
}

程式碼解析

obj->parent = malloc(n * sizeof(int *));

可以從這一行看出 parent 為 pointer to pointer 的 type

int max_level = 32 - __builtin_clz(n) + 1;
for (int i = 0; i < n; i++) {
    obj->parent[i] = malloc(max_level * sizeof(int));
    for (int j = 0; j < max_level; j++)
        obj->parent[i][j] = -1;
}

max_level是紀錄現在這個tree當中最大的level數為多少

將 obj->parent 做成二維陣列，obj->parent[n][max_level] 並將所有值初始化成 -1

( -1 表示 no ancestor 的意思)

for (int i = 0; i < parentSize; i++)
    obj->parent[i][0] = parent[i];

parent[i] 存放的是 node_i 父親的位置

並把obj->parent[i][0] 用來存放 node_i 的父親

for (int j = 1;; j++) {
        int quit = 1;
        for (int i = 0; i < parentSize; i++) {
            obj->parent[i][j] = obj->parent[i][j-1] == -1 ? -1:obj->parent[obj->parent[i][j - 1]];
            if (obj->parent[i][j] != -1) quit = 0;
        }
        if (quit) break;
}

目的: 設置 obj->parent 的 element

測驗3-FizzBuzz

從 1 數到 n

• 如果是 3的倍數，印出 “Fizz”
• 如果是 5 的倍數，印出 “Buzz”
• 如果是 15 的倍數，印出 “FizzBuzz”
• 如果都不是，就印出數字本身

可以觀察到:

• 整數格式字串 "%d" : 長度為 2
• "Fizz" 或 "Buzz" : 長度為 4
• "FizzBuzz" : 長度為 8

#define MSG_LEN 8

static inline bool is_divisible(uint32_t n, uint64_t M) {
    return n * M <= M - 1;
}

static uint64_t M3 = UINT64_C(0xFFFFFFFFFFFFFFFF) / 3 + 1;
static uint64_t M5 = UINT64_C(0xFFFFFFFFFFFFFFFF) / 5 + 1;

int main(int argc, char *argv[]) {
    for (size_t i = 1; i <= 100; i++) {
        uint8_t div3 = is_divisible(i, M3);
        uint8_t div5 = is_divisible(i, M5);
        unsigned int length = (2 << div3) << div5;

        char fmt[MSG_LEN + 1];
        strncpy(fmt, &"FizzBuzz%u"[(MSG_LEN >> div5) >> (div3 << 2)], length);
        fmt[length] = '\0';

        printf(fmt, i);
        printf("\n");
    }
    return 0;
}

程式解說

div3、div5 分別可知道 i 是否為 3 或 5 的倍數

• 是，為 1
• 否，則為 0

這個概念在quiz2有碰過

length則是根據我們觀察，判斷是不是3或5的倍數進行 operator 的平移

div3	div5	length	( 8 >> div5) >> (div3 << 2)
1	0	4	0
0	1	4	4
1	1	8	0
0	0	2	8