#include <iostream>
#include <cmath>
using namespace std;

int main() {
    int count = 0;  // 用于统计素数的个数
    // 遍历 100 到 200 之间的每一个数
    for (int num = 100; num <= 200; num++) {
        bool isPrime = true;  // 先假设当前数是素数
        // 判断是否为素数，检查从 2 到根号 num 的数
        for (int i = 2; i <= sqrt(num); i++) {
            if (num % i == 0) {
                isPrime = false;  // 如果能整除，则不是素数
                break;
            }
        }
        if (isPrime) {
            count++;  // 如果是素数，计数器加 1
            cout << num << " ";  // 输出素数
        }
    }
    cout << endl << count << endl;  // 换行后输出素数的个数
    return 0;
}

#include <iostream>
#include <vector>
#include <cmath>

// 计算两点间距离
double distance(std::vector<double>& a, std::vector<double>& b) {
    return std::sqrt((a[0] - b[0]) * (a[0] - b[0]) + (a[1] - b[1]) * (a[1] - b[1]));
}

// 计算三角形面积
double triangleArea(std::vector<double>& a, std::vector<double>& b, std::vector<double>& c) {
    double ab = distance(a, b);
    double bc = distance(b, c);
    double ca = distance(c, a);
    double s = (ab + bc + ca) / 2;
    return std::sqrt(s * (s - ab) * (s - bc) * (s - ca));
}

// 判断点是否在三角形内
bool isPointInTriangle(std::vector<double>& p, std::vector<double>& a, std::vector<double>& b, std::vector<double>& c) {
    double s = triangleArea(a, b, c);
    double s1 = triangleArea(p, a, b);
    double s2 = triangleArea(p, b, c);
    double s3 = triangleArea(p, c, a);
    // 考虑浮点数精度问题，判断差值是否在一个很小的范围内
    return std::abs(s - (s1 + s2 + s3)) < 1e-9;
}

int main() {
    // 测试代码
    std::vector<double> a = {0, 0};
    std::vector<double> b = {1, 0};
    std::vector<double> c = {0, 1};
    std::vector<double> p = {0.2, 0.2};
    std::cout << (isPointInTriangle(p, a, b, c)? "在三角形内" : "不在三角形内") << std::endl;
    return 0;
}

#include <iostream>
using namespace std;

// 判断是否能构成有效的鸡兔同笼问题
bool isSameCage(int heads, int feet) {
    // 判断脚的数量减去两倍头的数量是否为偶数且大于等于0
    if ((feet - 2 * heads) % 2 == 0 && feet - 2 * heads >= 0) {
        int y = (feet - 2 * heads) / 2;  // 兔子的数量
        int x = heads - y;  // 鸡的数量
        // 判断鸡和兔的数量是否都为非负整数
        if (x >= 0 && y >= 0) {
            return true;
        }
    }
    return false;
}
int main() {
    // 测试用例
    cout << (isSameCage(35, 94)? "true" : "false") << endl;
    cout << (isSameCage(2, 4)? "true" : "false") << endl;
    cout << (isSameCage(4, 2)? "true" : "false") << endl;
    cout << (isSameCage(10, 30)? "true" : "false") << endl;
    return 0;
}

#include <iostream>
#include <string>
using namespace std;

int main() {
    string s;
    cin >> s;
    for (char& c : s) {
        if (isupper(c)) {
            c = tolower(c);
        } else if (islower(c)) {
            c = toupper(c);
        }
    }
    cout << s << endl;
    return 0;
}

#include <iostream>
#include <string>
using namespace std;

int main() {
	string s;
	cin >> s;
	for (char& c : s) {
	if ('A'<=c && c<='Z') {
		c +=32;
	}
	else if ('a' <= c && c <= 'z') {
		c -=32;
	}
	}
	cout << s << endl;
    return 0;
}

#include<stdio.h>
#include<iostream>
#include<sstream>
#include<vector>
#include<string>
#include<algorithm>
using namespace std;

//判断是否为正方形：正方形的四条边长度相等，对角线长度相等，且对象线长度为边长的根号2倍
// 难点：四个点并不是按照顺序输入的。
//1、本题只计算 边长的平方，对角线的平方 ， 能给计算带来方便
//2、本题还有一个巧的地方：就是四个点中两两之间的线段刚好就是6条

int distanceSquare(int x1, int y1, int x2, int y2)
{
	return (y1 - y2) * (y1 - y2) + (x1 - x2) * (x1 - x2);
}

int main()
{
	vector<vector<int>> square(4, vector<int>(2, 0));
	for (int i = 0; i < 4; i++)
	{
		for (int j = 0; j < 2; j++)
		{
			cin >> square[i][j];
		}
	}
	vector<int> distance;
	distance.push_back(distanceSquare(square[0][0], square[0][1], square[1][0], square[1][1]));
	distance.push_back(distanceSquare(square[1][0], square[1][1], square[2][0], square[2][1]));
	distance.push_back(distanceSquare(square[2][0], square[2][1], square[3][0], square[3][1]));
	distance.push_back(distanceSquare(square[3][0], square[3][1], square[0][0], square[0][1]));
	distance.push_back(distanceSquare(square[0][0], square[0][1], square[2][0], square[2][1]));
	distance.push_back(distanceSquare(square[1][0], square[1][1], square[3][0], square[3][1]));

	sort(distance.begin(), distance.end());

	if (distance[0] == distance[1] && distance[1] == distance[2] && distance[2] == distance[3]
		&& distance[4] == distance[5] && distance[5] == 2 * distance[0])
	{
		cout << "Y";
	}
	else
	{
		cout << "N";
	}
	return 0;
}