Analysis of past CSP-J preliminary competition questions | 2021 CSP-J preliminary competition improvement procedures (39-43)

Learn C++ from a baby! Record the questions in the CSP-J exam preparation study process and record every moment.

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#include <iostream>

using namespace std;

struct point {
    int x, y, id;
};

bool equals(point a, point b) {
    return a.x == b.x && a.y == b.y;
}

bool cmp(point a, point b) {  //相当于a<b
    return ①;
}

void sort(point A[], int n) {  //将关键点的集合按照从小到大方式排序
    for ( int i=0; i<n; i++)
        for (int j=1; j<n; j++)
            if (cmp(A[j], A[j=1])) {
                point t = A[j];
                A[j] = A[j-1];
                A[j-1] = t;
            }
}

int unique(point A[], int n) {  //unique为去重
    int t = 0;
    for (int i=0; i<n; i++)
        if (②)
            A[t++] = A[i];
    return t;
}

bool binary_search(point A[], int n, int x, int y) {  //二分搜索，判断p是否在A[0]-A[n-1]中
    point p;
    p.x = x;
    p.y = y;
    p.id = n;
    int a = 0, b = n - 1;
    while (a<b) {
        int mid = ③;
        if (④)  
            a = mid + 1;
        else
            b = mid;
    }
    return equals(A[a], p);
}

const int MAXN = 1000;
point A[MAXN];

int main() {
    int n;
    cin >> n;
    for (int i=0; i<n; i++) {
        cin >> A[i].x >> A[i].y;
        A[i].id = i;
    }
    sort(A, n);
    n = unique(A,n);
    int ans = 0;
    for (int i=0; i<n; i++)
        for (int j=0; i<n; j++)
            if (⑤ && binary_search(A, n, A[i].x, A[j].y) && binary_search(A, n, A[j].x, A[i].y)) {
                ans ++;
            }
    cout << ans << endl;
    return 0;
}

39. ① should be filled in ( )

A.a.x != b.x ? a.x < b.x : a.id < b.id

B.a.x != b.x ? a.x < b.x : a.y < b.y

C.equals(a, b) ? a.id < b.id : a.x < b.x

D.equals(a, b) ? a.id < b.id : (a.x != b.x ? a.x < b.x : a.y < b.y)

[Answer]: B

【Analysis】

Because equals compares coordinates, and cmp can only compare coordinates, all options with ids cannot be selected, so you can only select B. The priority is to compare x first and then y.

40. ② should be filled in ( )

A.i == 0 || cmp(A[i], A[i-1])

B.t == 0 || equals(A[i], A[t-1])

C.i == 0 || !cmp(A[i], A[i-1])

D.t == 0 || !equals(A[i], A[t-1])

[Answer]: D

【Analysis】

When A[t-1] is not equal to A[i], put A[i] at the position of A[t], and then t++. Choose D

41. ③ should be filled in ( )

A.b - (b - a) / 2 +1

B.(a + b + 1) >> 1

C.(a + b) >> 1

D.a + (b - a + 1) / 2

[Answer]: C

【Analysis】

There are many binary search methods. There is no need to memorize them. The only criterion is that there cannot be an infinite loop. The infinite loop occurs when a and b are closest (the difference between a and b is 1), that is, the interval is [a, a+ 1]. At this time, it needs to be split into two intervals [a] and [a+1], and the intervals split into by binary search are [a, mid], [mid+1, a+1], then it is necessary to infer that the value of mid equal to a. You can substitute b=a+1, choose C. Shifting one position to the right is equivalent to dividing by 2.

42. ④ should be filled in ( )

A.!cmp(A[mid], p)

B.cmp(A[mid], p)

C.cmp(p, A[mid])

D.!cmp(p, A[mid])

[Answer]: B

【Analysis】

The purpose is to eliminate the part that is not equal to p, 4 options. Only when option B is true, A[0]...A[mid] cannot be equal to p, because here is A[0]...A[mid ] is less than p

43. ⑤ should be filled in ( )

A.A[i].x == A[j].x

B.A[i].id < A[j].id

C.A[i].x == A[j].x && A[i].id < A[j].id

D.A[i].x < A[j].x && A[i].y < A[j].y

[Answer]: D

【Analysis】

A[i] should be at the lower left of A[j]. The goal of the program is to find a point whose coordinates are equal to A[i].x and A[j].y between point i and point j through binary search.