使用C/C++两种语言完成50道题目，体会两种语言之间的不同。编译环境都是在VS2019，完成时间2020.06.21-2020.06.22 。由于C++ 对C的兼容性，部分main()函数没有写return 0 也可以执行成功，但是不提倡,最好加上，否则其他编译器可能报错。C语言不需要严格对齐，因此不对CSDN 的格式做过多的纠正。

31. 编一程序，输入一行字符串，统计其中的小写英文字母的个数

C

#include<stdio.h>
int main()
{char st[100];printf("请输入字符串：");scanf_s("%s", st, 100);int i = 0, num = 0;while (st[i] != '\0'){if (st[i] >= 'a' and st[i] <= 'z')num=num+1;i++;}printf("%d",num);
}

C++

#include<iostream>
using namespace std;
int main()
{char s[100];cin.getline(s,100);int i = 0, num = 0;while (s[i]!='\0'){if (s[i] >= 'a' and s[i] <= 'z')num++;i++;}cout << num;
}

32. 编一程序，输入一行字符串，将其中的大写英文字母改为小写，再输出

C

#include<stdio.h>
int main()
{char s[100];scanf_s("%s", s, 100);int i = 0, num = 0;while (s[i] != '\0'){if (s[i] >= 'A' and s[i] <= 'Z')s[i] += 32;i++;}i = 0;while (s[i]!='\0'){printf("%c", s[i]);i++;}
}

C++

#include<iostream>
using namespace std;
int main()
{char s[100];cin.getline(s, 100);int i = 0, num = 0;while (s[i] != '\0'){if (s[i] >= 'A' and s[i] <= 'Z')s[i]+=32;i++;}cout << s<<endl;
}

33. 打印杨辉三角形（帕斯卡三角形），打印10行

C

#include<stdio.h>
int main()
{int a[10][10] = { 0 };for (int i = 0; i < 10; i++){a[i][0] = 1;a[i][i] = 1;}for (int i = 1; i < 10; i++)for (int j = 1; j < i; j++)a[i][j] = a[i - 1][j - 1] + a[i - 1][j];for (int i = 0; i < 10; i++){for (int j = 0; j <= i; j++)printf("%d \t", a[i][j]);printf("\n");}
}

C++

#include<iostream>
#include<iomanip> 
using namespace std;
int main()
{int a[10][10] = { 0 };for (int i = 0; i < 10; i++){a[i][0] = 1;a[i][i] = 1;}for (int i = 1; i < 10; i++)for (int j = 1; j < i; j++)a[i][j] = a[i - 1][j - 1] + a[i - 1][j];for (int i = 0; i < 10; i++){for (int j = 0; j <= i; j++)cout << setw(4) << a[i][j];cout << endl;}
}

34. 打印一个九九乘法表

C

#include<stdio.h>
int main()
{int i, j;for (i = 1; i < 10; i++){for (j = 1; j <= i; j++){printf("i*j=%d\t", i * j);}printf("\n");}return 0;
}

C++

#include<iostream>
using namespace std;
int main()
{int i, j;for (i = 1; i < 10; i++) {for (j = 1; j <= i; j++){cout << i<<"*"<< "="<< i*j<< " ";}cout << endl;}
}

35. 掷骰子10000次，统计得到各点数的次数

C

#include<stdio.h>
#include<stdlib.h>
#include<time.h>
int main()
{int a[7] = { 0 };srand(time(0));for (int i = 1; i <= 10000; ++i)a[1 + rand() % 6]++;for (int i = 1; i <= 6; i++)printf("%d : %d\n", i, a[i]);
}

C++

#include<iostream>
using namespace std;
#include<stdlib.h>
#include<time.h>
int main()
{int a[7] = { 0 };srand(time(0));for (int i = 1; i <= 10000; ++i)a[1 + rand() % 6]++;for (int i = 1; i <= 6; ++i)cout << i << ": " << a[i] << endl;
}

36. 编写函数distance，计算两点(x1,y1)和(x2,y2)之间的距离

C

#include<stdio.h>
#include<math.h>
double dis(double, double, double, double);
int main()
{double(d);double x1 = 0, x2 = 4, y1 = 0, y2 = 3;d = dis(x1, x2, y1, y2);printf("%lf",d);
}
double dis(double x1, double x2, double y1, double y2)
{double d = sqrt((y1 - y2) * (y1 - y2) + (x1 - x2) * (x1 - x2));return d;
}

C++

#include<iostream>
using namespace std;
double dis(double, double,double, double);
int main()
{double(d);double x1 = 0, x2 = 4, y1 = 0, y2 = 3;d = dis(x1, x2, y1, y2);cout << d;
}
double dis(double x1, double x2, double y1,double y2)
{double d = sqrt((y1 - y2) * (y1 - y2) + (x1 - x2) * (x1 - x2));return d;
}

37. 写一个程序，进行体操评分，依次输入10名评委所评分数，去除一个最高分和一个最低分，再算出平均分作为选手的得分

C

#include<stdio.h>
int main()
{int a, max1, min1;float total, ave;printf("请输入十位评委的分数");scanf_s("%d",&a);max1 = min1 = total = a;for (int i = 0; i < 9; i++){scanf_s("%d", &a);if (max1 < a) max1 = a;if (min1 > a)  min1 = a;total = total + a;}total = total - min1 - max1;ave = total / 8;printf( "%lf",ave);return 0;
}

C++

#include<iostream>
using namespace std;
int main()
{float a,max, min, total, ave;cout << "请输入十位评委的分数" << endl;cin >> a;max = min =total= a;for (int i=0;i<9;i++){cin >> a;if (max < a) max = a;if (min > a)  min = a;total = total + a;}total = total - min - max;ave = total / 8;cout << ave;
}

38. 写一函数，将一数组中的元素反转

C

#include<stdio.h>
void  reverse(int*, int);
int main()
{int a[10] = { 1,2,3,4,5,6,7,8,9,10 };int n = sizeof(a) / sizeof(a[0]);reverse(a, n);for (int i = 0; i < 10; i++)printf("%d ",a[i]);
}
void  reverse(int* a, int n)
{int temp;for (int i = 0; i < n / 2; i++){temp = a[i];a[i] = a[n - 1 - i];a[n - 1 - i] = temp;}
}

C++

#include<iostream>
using namespace std;
void  reverse(int *,int);
int main()
{int a[10] = { 1,2,3,4,5,6,7,8,9,10 };int n = sizeof(a)/sizeof(a[0]);reverse(a,n);for (int i = 0; i < 10; i++)cout << a[i]<<" ";
}
void  reverse(int* a,int n)
{int temp;for (int i = 0; i < n / 2; i++){temp = a[i];a[i] = a[n - 1 - i];a[n - 1 - i] = temp;}
}

39. 写一函数，在一个数组中找出最大元素的位置

C++

#include<iostream>
using namespace std;
void  findmax(int*, int);
int main()
{int a[10] = { 1,2,3,4,25,16,97,8,9,10 };int n = sizeof(a) / sizeof(a[0]);findmax(a, n);
}
void findmax(int * a,int n)
{int max = a[0];int k=0;for (int i = 1; i < n; i++){if (a[i] > max) {max = a[i];k = i;}}cout << "位置在第" << k + 1<<"个";
}

40. 找出一个二维数组中的鞍点，即该元素在该行上最大，在该列上最小

C

#include <stdio.h>
#include <stdlib.h>
int main()
{int a[3][3];int i, j, k, p, q, max, min;for (i = 0; i < 3; i++) {for (j = 0; j < 3; j++)scanf_s("%d", &a[i][j]);               //输入数组}for (i = 0; i < 3; i++) {for (j = 0, max = 0, p = 0; j < 3; j++) {         //判断出该行元素的最大值if (a[i][j] > max) {max = a[i][j];p = j;}else {max = max;p = p;}}for (k = 0, min = a[i][p], q = i; k < 3; k++) {   //判断最大值在该列是否为最小值if (a[k][p] < min) {min = a[k][p];q = k;}else {min = min;q = q;}}if (i == q) {printf("%d is the saddle point!\n", a[i][p]);  //如果该最大值行号没有改变，输出鞍点break;}}if (i == 3 && i != q)printf("There is no saddle point!\n");             //输出没有鞍点system("pause");return 0;
}

C++

#include<iostream>
using namespace std;
int main()
{int a[3][3];int i, j, k, p, q, max, min;for (i = 0; i < 3; i++) {for (j = 0; j < 3; j++)cin>>a[i][j];               //输入数组}for (i = 0; i < 3; i++) {for (j = 0, max = 0, p = 0; j < 3; j++) {         //判断出该行元素的最大值if (a[i][j] > max) {max = a[i][j];p = j;}else {max = max;p = p;}}for (k = 0, min = a[i][p], q = i; k < 3; k++) {   //判断最大值在该列是否为最小值if (a[k][p] < min) {min = a[k][p];q = k;}else {min = min;q = q;}}if (i == q) {cout << a[i][p] <<" is the saddle point!";  //如果该最大值行号没有改变，输出鞍点break;}}if (i == 3 && i != q)cout<<"There is no saddle point!"<<endl;             //输出没有鞍点
}