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Homework Exercises 1

news 来源：原创 2024/5/12 6:34:22

第一题

返回所有这样的整数k的和：

1 <= k < N
N可被k整除

/* 1a. */
/** Returns the sum of all integers, k, such that 1 <= k < N and
 *  N is evenly divisible by k. */
static int factorSum(int N) {
    // Fill in here
}

static int factorSum(int N) {
    int i = 0;
    int j = 0;
    for (; i <= N / 2; i++) {
        if (N % i == 0) {
            j += i;
        }
    }
    return j;
}

注意无需遍历N，只需要到N/2就行了。

打印出0 - N间的所有相亲数，每一对占一行，小的在前，大的在后。

所谓想亲数就是：指两个正整数中，彼此的全部约数之和（本身除外）与另一方相等。

例如220与284：

220的全部约数(除掉本身)相加是：1+2+4+5+10+11+20+22+44+55+110=284
284的全部约数(除掉284本身)相加的和是：1+2+4+71+142=220

/* 1b. */
/** Print the set of all sociable pairs whose members are all
 *  between 1 and N>=0 (inclusive) on the standard output (one pair per
 *  line, smallest member of each pair first, with no repetitions). */
static void printSociablePairs(int N) {
    // Fill in here
}

static void printSociablePairs(int N) {
    int i = 0;
    int j = 0;
    for(; i<=N; i++){
        j = factorSum(i)
        if (j > i){
            system.out.println(i + ", " + j);
        }
    }
}

第二题

根据下面的简单链表实现完成下面题目。

import java.util.Formatter;

/** Scheme-like pairs that can be used to form a list of integers.
 *  @author P. N. Hilfinger
 *  [Do not modify this file.]
 */
public class IntList {
    /** First element of list. */
    public int head;
    /** Remaining elements of list. */
    public IntList tail;

    /** A List with head HEAD0 and tail TAIL0. */
    public IntList(int head0, IntList tail0) {
        head = head0;
        tail = tail0;
    }

    /** A List with null tail, and head = 0. */
    public IntList() {
        /* NOTE: public IntList () { }  would also work. */
        this (0, null);
    }

    /** Returns a new IntList containing the ints in ARGS. */
    public static IntList list(Integer ... args) {
        IntList result, p;

        if (args.length > 0) {
            result = new IntList(args[0], null);
        } else {
            return null;
        }

        int k;
        for (k = 1, p = result; k < args.length; k += 1, p = p.tail) {
            p.tail = new IntList(args[k], null);
        }
        return result;
    }

    /** Returns true iff X is an IntList containing the same sequence of ints
     *  as THIS. */
    public boolean equals(Object x) {
        if (!(x instanceof IntList)) {
            return false;
        }
        IntList L = (IntList) x;
        IntList p;
        for (p = this; p != null && L != null; p = p.tail, L = L.tail) {
            if (p.head != L.head) {
                return false;
            }
        }
        if (p != null || L != null) {
            return false;
        }
        return true;
    }

    @Override
    public int hashCode() {
        return head;
    }

    @Override
    public String toString() {
        Formatter out = new Formatter();
        String sep;
        sep = "(";
        for (IntList p = this; p != null; p = p.tail) {
            out.format("%s%d", sep, p.head);
            sep = ", ";
        }
        out.format(")");
        return out.toString();
    }

}

返回一个IntList包含IntList A，后面跟着IntList B，不允许使用new。

/* 2a. */
/** Returns a list consisting of the elements of A followed by the
 *  elements of B.  May modify items of A. Don't use 'new'. */
static IntList dcatenate(IntList A, IntList B) {
    // Fill in here
}

static IntList dcatenate(IntList A, IntList B) {
    IntList p
    for (p = A; p != null; p = p.tail;){
        if(p.tail == null){
            p.tail = B;
        }
    }
    
    return A;
}

注意链表的概念。

返回一个IntList的一部分，从原IntList L的第start个元素开始，长度为len。不允许改变原IntList的每个元素，如果元素不存在会抛出错误。

/* 2b. */
/** Returns the sublist consisting of LEN items from list L,
 *  beginning with item #START (where the first item is #0).
 *  Does not modify the original list elements.
 *  It is an error if the desired items don't exist. */
static IntList sublist(IntList L, int start, int len) {
    //Fill in here
}

static IntList sublist(IntList L, int start, int len) {
    IntList p;
    IntList k;
    IntList j;
    int i = 0;
    int head0;
    for (p = L; i < start + len; p = p.tail, i++){
        if (i == start){
            head0 = p.head;
            k = new IntList(head0, null);
            j = k;
        } else if (i > start && i < start + len)
            head0 = p.head;
            j.tail = new IntList(head0, null);
            j = j.tail;
        }
    }
        
    return k;
}

返回一个IntList的一部分，从原IntList L的第start个元素开始，长度为len。需要改变原IntList的元素，如果元素不存在会抛出错误。

/* 2c. */
/** Returns the sublist consisting of LEN items from list L,
 *  beginning with item #START (where the first item is #0).
 *  May modify the original list elements. Don't use 'new'.
 *  It is an error if the desired items don't exist. */
static IntList dsublist(IntList L, int start, int len) {
    // Fill in here
}

static IntList dsublist(IntList L, int start, int len) {
    IntList p;
    int i = 0;
    for (p = L; i <= start + len; p = p.tail, i++){
        if (i == start){
            L = p;
        } else if (i == start + len){
            p = null;
        }
    }
        
    return L;
}