选票系统

最新推荐文章于 2024-10-06 08:50:50 发布

原创最新推荐文章于 2024-10-06 08:50:50 发布 · 795 阅读

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#C++

该博客介绍了如何使用C++编程实现一个选票系统，包括矩阵转置操作以统计字符类别的频率，并探讨了利用递归进行数据处理，如分鱼的方法。

//选票系统
#include<iostream>
using namespace std;
int main()
{
	int i, v0 = 0, v1 = 0, v2 = 0, v3 = 0, n, a[50];
	cout << "Please enter the number of elestorrate:";
	cin >> n;
	cout << "Please enter 1 or 2 or 3 ";
	for (i = 0; i < n; i++)
		cin >> a[i];
	for (i = 0; i < n; i++)
	{
		if (a[i] ==1)
			v1++;
		if (a[i] == 2)
			v2++;
		if (a[i] == 3)
			v3++;
		else
			v0++;
	}
	cout << "candiate 1:" << v1 << endl;
	cout << "candiate 2:" << v2 << endl;
	cout << "candiate 3:" << v3 << endl;
	cout << "not use:" << v0<< endl;
	return 0;
}

矩阵转置

//矩阵转置
#include<iostream>
using namespace std;
int main()
{
	int i, j, i1, j1, a[101][101], b[101][101];
	cout << "please enter the number of rows(<=100)";
	cin >> i1;
	cout << "please enter the number of columns(<=100)";
	cin >> j1;
	for (i = 1; i <= i1; i++)
	for (j = 1; j <= j1; j++)
		cin >> a[i][j];
	cout << "a[i][j]= " << endl;
	for (i = 1; i <= i1; i++)
	{
		for (j = 1; j <= j1; j++)
		{
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
	cout << "b[i][j]= " << endl;
	for (j = 1; j<= j1; j++)
	{
		for (i = 1;i <= i1; i++)
		{
			cout << a[i][j] << " ";
		}
		cout << endl;
	}
	return 0;
}

统计字符类别

//main program
#include<iostream>
using namespace std;
int main()
{
	char c[5];
	int i;
	int letters = 0, space = 0, digit = 0, others = 0;
	cout << "Please input some characters:" << endl;
	for (i = 0; i < 5; i++)
		cin >> c[i];
	for (i = 0; i < 5;i++)
	{
		if (c[i] >= 'a'&& c[i] <= 'z' || c[i] >= 'A'&& c[i] <= 'Z')
			letters++;
		else if (c[i] == ' ')
			space++;
		else if (c[i] >= '0'&& c[i] <= '9')
			digit++;
		else
			others++;
	}
	cout << "char= " << letters << endl;
	cout << "space= " << space << endl;
	cout << "digit= " << digit << endl;
	cout << "others= " << others << endl;
}

递归分鱼

//递归解决分鱼问题
#include<iostream>
using namespace std;
int yu(int i, int num);
int main()
{
	int n = 1,sum;
	int num;
	cout << "please enter the number of fish:"<<endl;
	cin >> num;
	do
	{
		sum = yu(5,num);
		num++;
	} while (sum%1 != 0);
	cout << "the total fish is " << sum << endl;
	return 0;

}
int yu(int i,int num)
{
	if (i == 1)
		return (5*num + 1);
	else
		return (5*yu(i - 1,num) + 1);
}