C++ 综合复习与实战

#include <iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    int a[100];
    int sum = 0, mx = -1, mn = 101, pass = 0;
    for (int i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
        if (a[i] > mx) mx = a[i];
        if (a[i] < mn) mn = a[i];
        if (a[i] >= 60) pass++;
    }
    cout << "平均: " << (double)sum/n << endl;
    cout << "最高: " << mx << endl;
    cout << "最低: " << mn << endl;
    cout << "及格: " << pass << endl;
    return 0;
}

#include <iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    int mid = n / 2;
    // 上半部分（含中间行）
    for (int i = 0; i <= mid; i++) {
        for (int j = 0; j < mid - i; j++) cout << " ";
        for (int j = 0; j < 2*i+1; j++) cout << "*";
        cout << endl;
    }
    // 下半部分
    for (int i = mid-1; i >= 0; i--) {
        for (int j = 0; j < mid - i; j++) cout << " ";
        for (int j = 0; j < 2*i+1; j++) cout << "*";
        cout << endl;
    }
    return 0;
}

#include <iostream>
using namespace std;
int main() {
    int n;
    cin >> n;
    int a[100][100];
    for (int i = 0; i < n; i++)
        for (int j = 0; j < n; j++)
            cin >> a[i][j];

int sum = 0;
    for (int i = 0; i < n; i++) {
        sum += a[i][i];           // 主对角线
        if (i != n-1-i)         // 避免中心元素重复
            sum += a[i][n-1-i];   // 副对角线
    }
    cout << sum << endl;
    return 0;
}

int n, sum = 0;
cin >> n;
for (int i = 1; i <= n; i++) sum += i;
cout << sum << endl;

int a[10], mx = -1000000;
for (int i = 0; i < 10; i++) {
    cin >> a[i];
    if (a[i] > mx) mx = a[i];
}
cout << mx << endl;

int n, rev = 0;
cin >> n;
while (n > 0) {
    rev = rev * 10 + n % 10;
    n /= 10;
}
cout << rev << endl;

int a[100][100], b[100][100], c[100][100];
for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
        cin >> a[i][j];
for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
        cin >> b[i][j];
for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++)
        c[i][j] = a[i][j] + b[i][j];

for (int i = 0; i < n-1; i++) {
    int minPos = i;
    for (int j = i+1; j < n; j++)
        if (a[j] < a[minPos]) minPos = j;
    if (minPos != i) {
        int t = a[i]; a[i] = a[minPos]; a[minPos] = t;
    }
}

// C[i][j] = sum(A[i][k] * B[k][j]) for k=0..m-1
for (int i = 0; i < n; i++)
    for (int j = 0; j < p; j++) {
        c[i][j] = 0;
        for (int k = 0; k < m; k++)
            c[i][j] += a[i][k] * b[k][j];
    }

// 利用组合数公式 C(n,k) = n!/(k!*(n-k)!)
// 或用递推：C(n,k) = C(n,k-1) * (n-k+1) / k
long long c = 1;
cout << c << " ";
for (int k = 1; k <= n; k++) {
    c = c * (n - k + 1) / k;
    cout << c << " ";
}

// 用结构体存储姓名和生日，用冒泡或选择排序
struct Student {
    string name;
    int y, m, d;
} s[100];
// 比较函数：先比年，再比月，再比日
for (int i = 0; i < n-1; i++)
    for (int j = 0; j < n-1-i; j++)
        if (s[j].y > s[j+1].y ||
            (s[j].y == s[j+1].y && s[j].m > s[j+1].m) ||
            (s[j].y == s[j+1].y && s[j].m == s[j+1].m && s[j].d > s[j+1].d))
            swap(s[j], s[j+1]);

// 与案例5螺旋矩阵相同，此处不再重复
// 核心：四个边界变量控制填充方向

// 分别检查行、列、宫格
bool checkRow() {
    for (int i = 0; i < 9; i++) {
        bool vis[10] = {0};
        for (int j = 0; j < 9; j++)
            if (vis[a[i][j]]) return false;
            else vis[a[i][j]] = true;
    }
    return true;
}
// 列和宫格同理，宫格用 (i/3*3+j/3) 定位

// Kadane算法：O(n) 时间复杂度
int maxSum = a[0], cur = a[0];
for (int i = 1; i < n; i++) {
    cur = max(a[i], cur + a[i]);
    maxSum = max(maxSum, cur);
}
cout << maxSum << endl;

int a[100] = {0}; // 0表示在圈内，1表示已出列
int cnt = 0, pos = 0, remain = n;
while (remain > 0) {
    if (a[pos] == 0) cnt++;
    if (cnt == m) {
        cout << pos+1 << " ";
        a[pos] = 1;
        cnt = 0;
        remain--;
    }
    pos = (pos + 1) % n; // 环形
}