题目编号
语言
全部语言
等级
全部等级
知识点
选择知识点 (0)
找到 1260 道单选题
EXY-SC-1140
第 361 题

下面 fibonacci 函数的时间复杂度为( )。

int fibonacci(int n) {
    if (n <= 1)
        return n;
    else
        return fibonacci(n - 1) + fibonacci(n - 2);
}

(注:本题GESP官方给的选项B是 $O(\phi^n), \phi = \frac{\sqrt{5}-1}{2}$)

A

$O(1)$

B

$O(\phi^n), \phi = \frac{\sqrt{5}+1}{2}$

C

$O(n)$

D

$O(n \log n)$

语言: C++
GESP真题 八级
2024.9
单选题号: 15
EXY-SC-1139
第 362 题

下面程序的 Merge_Sort 函数时间复杂度为( )。

void Merge(int a[], int left, int mid, int right) {
    int temp[right - left + 1];
    int i = left;
    int j = mid + 1;
    int k = 0;
    while (i <= mid && j <= right) {
        if (a[i] < a[j])
            temp[k++] = a[i++];
        else
            temp[k++] = a[j++];
    }
    while (i <= mid)
        temp[k++] = a[i++];
    while (j <= right)
        temp[k++] = a[j++];
    for (int m = left, n = 0; m <= right; m++, n++)
        a[m] = temp[n];
}
 
void Merge_Sort(int a[], int left, int right) {
    if (left == right)
        return;
    int mid = (left + right) / 2;
    Merge_Sort(a, left, mid);
    Merge_Sort(a, mid + 1, right);
    Merge(a, left, mid, right);
}
A

$O(n \log n)$

B

$O(n^2)$

C

$O(2^n)$

D

$O(\log n)$

语言: C++
GESP真题 八级
2024.9
单选题号: 14
EXY-SC-1138
第 363 题

下面 Floyd 算法中,横线处应该填入的是( )。

#include <iostream>
using namespace std;
 
#define N 21
#define INF 9999999
int map[N][N];
 
int main() {
    int n, m, t1, t2, t3;
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            if (i == j)
                map[i][j] = 0;
            else
                map[i][j] = INF;
        }
    }
 
    for (int i = 1; i <= m; i++) {
        cin >> t1 >> t2 >> t3;
        map[t1][t2] = t3;
    }
 
    for (int k = 1; k <= n; k++)
        for (int i = 1; i <= n; i++)
            for (int j = 1; j <= n; j++)
                if (________) // 在此处填入选项
                    map[i][j] = map[i][k] + map[k][j];
 
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
            cout.width(4);
            cout << map[i][j];
        }
        cout << endl;
    }
}
A

map[i][j] < map[i][k] + map[k][j]

B

map[i][j] > map[i][k] + map[k][j]

C

map[i][j] > map[i][k] - map[k][j]

D

map[i][j] < map[i][k] - map[k][j]

语言: C++
GESP真题 八级
2024.9
单选题号: 13
EXY-SC-1137
第 364 题

下列 Dijkstra 算法中,横线处应该填入的是( )。

#include <iostream>
using namespace std;
 
#define N 100
int n, e, s;
const int inf = 0x7fffffff;
int dis[N + 1];
int cheak[N + 1];
int graph[N + 1][N + 1];
 
int main() {
    for (int i = 1; i <= N; i++)
        dis[i] = inf;
    cin >> n >> e;
    for (int i = 1; i <= e; i++) {
        int a, b, c;
        cin >> a >> b >> c;
        graph[a][b] = c;
    }
    cin >> s;
    dis[s] = 0;
    for (int i = 1; i <= n; i++) {
        int minn = inf, minx;
        for (int j = 1; j <= n; j++) {
            if (________) { // 在此处填入选项
                minn = dis[j];
                minx = j;
            }
        }
        cheak[minx] = 1;
        for (int j = 1; j <= n; j++) {
            if (graph[minx][j] > 0) {
                if (minn + graph[minx][j] < dis[j]) {
                    dis[j] = minn + graph[minx][j];
                }
            }
        }
    }
}
A

dis[j] > minn && cheak[j] == 0

B

dis[j] < minn && cheak[j] == 0

C

dis[j] >= minn && cheak[j] == 0

D

dis[j] < minn && cheak[j] != 0

语言: C++
GESP真题 八级
2024.9
单选题号: 12
EXY-SC-1136
第 365 题

下面 Prim 算法程序中,横线处应该填入的是()。

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;
int prim(vector<vector<int>> & graph, int n) {
    vector<int> key(n, INT_MAX);
    vector<int> parent(n, -1);
 
    key[0] = 0;
    for (int i = 0; i < n; i++) {
        int u = min_element(key.begin(), key.end()) - key.begin();
        if (key[u] == INT_MAX)
            break;
        for (int v = 0; v < n; v++) {
            if (__________) { // 在此处填入选项
                key[v] = graph[u][v];
                parent[v] = u;
            }
        }
    }
    int sum = 0;
    for (int i = 0; i < n; i++) {
        if (parent[i] != -1) {
            cout << "Edge: " << parent[i] << " - " << i << " Weight: " << key[i] << endl;
            sum += key[i];
        }
    }
    return sum;
}
int main() {
    int n, m;
    cin >> n >> m;
    vector<vector<int>> graph(n, vector<int>(n, 0));
    for (int i = 0; i < m; i++) {
        int u, v, w;
        cin >> u >> v >> w;
        graph[u][v] = w;
        graph[v][u] = w;
    }
    int result = prim(graph, n);
    cout << "Total weight of the minimum spanning tree: " << result << endl;
    return 0;
}
A

graph[u][v] >= 0 && key[v] > graph[u][v]

B

graph[u][v] <= 0 && key[v] > graph[u][v]

C

graph[u][v] == 0 && key[v] > graph[u][v]

D

graph[u][v] != 0 && key[v] > graph[u][v]

语言: C++
GESP真题 八级
2024.9
单选题号: 11
当前页显示 361 - 365 ,共 1260 道单选题