Analysis of the real questions of the CSP-J preliminary competition over the years | 2022 CSP-J preliminary competition reading program (28-34)

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#include <iostream>

using namespace std;

int n, k;

int solve1()  //返回使得x²≤n的最大整数x
{
    int l = 0, r = n;
    while (l<=r) {
        int mid = (l+r)/2;
        if (mid * mid <= n) l = mid+1;
        else r = mid -1;
    }
    return l -1;  //l-1是满足(l-1)²≤n的最大数
}

double solve2(double x)
{
    if (x == 0) return x;
    for (int i=0; i<k; i++)
        x = (x + n/x) / 2;  //假设x数列有极限A，随着k增加，xi会越来越接近√n
    return x;
}

int main()
{
    cin >> n >> k;
    double ans = solve2(solve1());
    cout << ans << ' ' << (ans * ans == n) << endl;
    return 0;
}

Assuming that int is a 32-bit signed integer type, the input n is a natural number not exceeding 47000, and k is a natural number not exceeding the range represented by int, complete the following judgment questions and multiple choice questions:

28. The most accurate time complexity analysis result of this algorithm is O(logn+k). ( )

[Answer]: yes

【Analysis】

The time complexity of solve1 is logn, and the time complexity of solve2 is k

29. When the input is "9801 1", the first number output is "99". ( )

[Answer]: yes

【Analysis】

The function of sovle1 is to find the square root of n. When n=9801, the square root of n is 99

30. For any input n, as the input k increases, the second output number will become "1". ( )

[Answer]: wrong

【Analysis】

If n is not the square of an integer, ans will be close to √n but not equal to √n due to round-off errors. The reason is that the double type will have a truncation error during the operation. For example, input "3 10", loop to the back, x is unchanged

31. The program has flaws. When the input n is too large, the multiplication in line 12 may overflow, so mid should be converted to a 64-bit integer before calculation. ( )

[Answer]: wrong

【Analysis】

The maximum value is (47000/2)^2, which does not exceed the maximum value of int 2147483647

32. When the input is "2 1", the first number output is closest to ( ).

A.1

B.1.414

C.1.5

D.2

[Answer]: C

【Analysis】

Substituting into the calculation, we get 1.5

33. When the input is "3 10", the first number output is closest to ( ).

A.1.7

B.1.732

C.1.75

D.2

[Answer]: B

【Analysis】

Substitute into the calculation to get √3. After several cycles of solve2, you will find that the value of x no longer changes

34. When the input is "256 11", the first number to output ( ).

A. is equal to 16

B. Close to but less than 16

C. Close to but greater than 16

D. The first three situations are all possible

[Answer]: A

【Analysis】

Same as question 29, the result is equal to 16