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  • C++ 考级试题

    1. 输出保留指定小数的浮点数: cout.precision()

     1 #include "iostream"
     2 using namespace std;
     3 int main()
     4 {
     5     double num;
     6     cin >> num;
     7     
     8     //
     9     cout.precision(12);
    10     cout << fixed << num;
    11     
    12     return 0;
    13 }
    View Code

    2. 设置输出数据的宽度:  cout.width()

     1 #include <iostream>
     2 using namespace std;
     3 
     4 int main()
     5 {
     6     for (int i = 0; i < 100; i++) {
     7         cout.width(3);                                  // 仅对下一个cout输出生效
     8         cout << i << endl;
     9     }
    10 }
    View Code

    3. 计算经过多少天之后是星期几

    1 // base:    当前是星期几
    2 // passed:  过了多少天
    3 int week(int base, int passed)
    4 {
    5     int week = (base+passed) % 7;
    6     int week_a[7] = {7,1,2,3,4,5,6};
    7     return week_a[week];
    8 }
    View Code

    4. Hello, World!的大小(sizeof)

     1 #include <iostream>
     2 using namespace std;
     3 
     4 int main(void)
     5 {
     6     int n;
     7     string hello = "Hello, World!";
     8     
     9     n = sizeof(hello);
    10     cout << n << endl;
    11     return 0;
    12 }
    View Code

     5. 创建素数表的函数

     1 #include <iostream>
     2 #include <vector>
     3 using namespace std;
     4 
     5 int creat_prime(int n, int prime[])
     6 {
     7     prime[0] = 0;
     8     prime[1] = 0;
     9     // 素数标志
    10     for (int i=2; i<n; i++) {
    11         prime[i] = 1;
    12     }
    13     // 遍历到根号n
    14     for (int i=2; (i*i)<n; i++) {
    15         if (prime[i] == 1) {
    16             for (int j=2*i; j<=n; j++) {
    17                 if ((j%i) == 0) {
    18                     prime[j] = 0;
    19                 }
    20             }
    21         }
    22     }
    23     return 0;
    24 }
    25 
    26 int main(void)
    27 {
    28     int n;
    29     cin >> n;
    30     //
    31     int *prime = new int[n];
    32     creat_prime(n, prime);
    33     for (int i=0; i<n; i++) {
    34         if (prime[i] != 0) {
    35             cout << i << endl;
    36         }
    37     }
    38     return 0;
    39 }
    View Code

     6. 辗转法求最大公约数

     1 #include <iostream>
     2 #include <vector>
     3 using namespace std;
     4 
     5 int gcd(int m, int n)
     6 {
     7     int r = m % n;
     8     if (r == 0) {
     9         return n;
    10     }
    11     else {
    12         return gcd(n, r);
    13     }
    14 }
    15 
    16 int main()
    17 {
    18     int a, b;
    19     cin >> a >> b;
    20     cout << gcd(a, b);
    21 }
    View Code
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  • 原文地址:https://www.cnblogs.com/linux-farmer/p/13638086.html
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