函数与结构体（入门6）

【深基7.例1】距离函数

#include <iostream>
#include <iomanip>
#include <cmath>
using namespace std;
int main()
{
	double x1, x2, x3, y1, y2, y3;
	cin >> x1 >> y1 >> x2 >> y2 >> x3 >> y3;

	double d1 = pow(pow(x2 - x1, 2) + pow(y2 - y1, 2), 0.5);
	double d2 = pow(pow(x3 - x1, 2) + pow(y3 - y1, 2), 0.5);
	double d3 = pow(pow(x3 - x2, 2) + pow(y3 - y2, 2), 0.5);

	cout <<fixed<< setprecision(2) << d1 + d2 + d3 << endl;

	return 0;
}

注意：

头文件cmath（pow的头文件）以及输出两位小数的方式

【深基7.例2】质数筛

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

bool fun(int x)
{
	if (x == 0 || x == 1)return false;
	if (x == 2)return true;
	if (x % 2 == 0)return false;
	else {
		for (int i = 3; i <= sqrt(x); i++) {
			if (x % i == 0)return false;
		}
	}
	return true;
}

int main()
{
	int n;
	cin >> n;
	vector<int>vec(n);
	for (int i = 0; i < n; i++) {
		cin >> vec[i];
	}
	bool flag= true;
	for (int i = 0; i < n; i++) {
		if (fun(vec[i])) {
			cout << vec[i] << ' ';
		}
	}


	return 0;
}

注意：

在判断是否为素数的函数中，for循环内应该是i<=sqrt(x)

【深基7.例3】闰年展示

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

bool solve(int x)
{
	if (x % 100 == 0) {
		if (x % 400 == 0)return true;
		else return false;
	}
	else {
		if (x % 4 == 0)return true;
		else return false;
	}
}

int main()
{
	int m, n;
	cin >> m >> n;
	vector<int>vec;
	int count = 0;
	for (int i = m; i <= n; i++) {
		if (solve(i)) {
			count++;
			vec.push_back(i);
		}
	}
	cout << count << endl;
	for (int i = 0; i < vec.size(); i++) {
		cout << vec[i] << " ";
	}

	return 0;
}

【深基7.例4】歌唱比赛

#include <iostream>
#include <vector>
#include <iomanip>
using namespace std;


int main()
{
	int m, n;
	cin >> n >> m;
	vector<int>vec;
	for (int i = 0; i < n; i++) {
		int maxn = -1;
		int minn = 11;
		int x;
		int sum = 0;
		for (int j = 0; j < m; j++) {
			cin >> x;
			if (maxn < x)maxn = x;
			if (minn > x)minn = x;
			sum += x;
		}
		sum = sum - minn - maxn;
		vec.push_back(sum);
	}
	int maxn = -1;
	for (int i = 0; i < vec.size(); i++) {
		if (vec[i] > maxn)maxn = vec[i];
	}
	cout << fixed << setprecision(2) << 1.0 * maxn / (m - 2) << endl;

	return 0;
}

【深基7.例7】计算阶乘

#include <iostream>
#include <vector>
#include <iomanip>
using namespace std;

long long  sum = 1;
int n;
void solve(int x)
{
	sum *= x;
	if (x == n)return;
	solve(++x);
}

int main()
{
	cin >> n;
	solve(1);
	cout << sum << endl;


	return 0;
}

注意：

在递归的时候，sum*=x应该写在判断x是否等于n之前，感觉是我递归不熟练，导致了错误

赦免战俘

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

bool a[2000][2000];

void dfs(int x, int y, int n)
{
	if (n) {
		for (int i = x; i < n / 2 + x; i++) {
			for (int j = y; j < n / 2 + y; j++) {
				a[i][j] = 0;
			}
		}
		dfs(n / 2 + x, y, n / 2);//右上角
		dfs(x, n / 2 + y, n / 2);//左下角
		dfs(n / 2 + x, n / 2 + y, n / 2);//右下角
	}
}

int main()
{
	int n;
	cin >> n;
	n = 1 << n;
	memset(a, true, sizeof(a));
	dfs(0, 0, n);
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < n; j++) {
			cout << a[i][j] << " ";
		}
		cout << endl;
	}

	return 0;
}

注意：

这里我借鉴的别人代码，由于本人对指针实在是不太了解，所以没有用大佬方法的指针，不过发现不用指针也可以做；

主要是模拟，然后递归遍历右上角，左下角，右下角；此外，数组a设置成了int类型的话，然后memset（a,1,sizeof(a)），不知道为什么在输出的时候出现了乱码，我现在还无法解释；后面把它设置成了bool类型，memset（a,true,sizeof(a)）就没问题。

【深基7.例9】最厉害的学生

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

struct score
{
	string name;
	int c;
	int m;
	int e;
	int sum;
}a[1005];

int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i].name >> a[i].c >> a[i].m >> a[i].e;
		a[i].sum = a[i].c + a[i].m + a[i].e;
	}
	int maxn = 0;
	int index = 0;
	for (int i = 0; i < n; i++) {
		if (a[i].sum > maxn) {
			maxn = a[i].sum;
			index = i;
		}
	}
	cout << a[index].name << " " << a[index].c << " " << a[index].m << " " << a[index].e << endl;

	return 0;
}

【深基7.例10】旗鼓相当的对手 - 加强版

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

struct score
{
	string name;
	int c;
	int m;
	int e;
	int sum;
}a[1005];

int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i].name >> a[i].c >> a[i].m >> a[i].e;
		a[i].sum = a[i].c + a[i].m + a[i].e;
	}
	for (int i = 0; i < n; i++) {
		for (int j = i + 1; j < n; j++) {
			if (abs(a[i].sum - a[j].sum) <= 10
				&& abs(a[i].m - a[j].m) <= 5
				&& abs(a[i].c - a[j].c) <= 5
				&& abs(a[i].e - a[j].e) <= 5) {
				cout << a[i].name << " " << a[j].name << endl;
			}
		}
	}

	return 0;
}

【深基7.例11】评等级

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

struct score
{
	int num;
	int score;
	int expand;
	int sum;

	void solve() {
		if (this->expand + this->score > 140 && this->sum >= 800) {
			cout << "Excellent" << endl;
		}
		else {
			cout << "Not excellent" << endl;
		}
	}
}a[1005];

int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i].num >> a[i].score >> a[i].expand ;
		a[i].sum = 7*a[i].score + 3*a[i].expand ;
	}
	for (int i = 0; i < n; i++) {
		a[i].solve();
	}

	return 0;
}

[NOIP2012 普及组] 质因数分解

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

bool isPrime(int x)
{
	if (x < 2)return false;
	if (x == 2)return true;
	for (int i = 2; i <= sqrt(x); i++) {
		if (x % i == 0)return false;
	}
	return true;
}

int main()
{
	int n;
	cin >> n;
	for (int i = 2; i < n; i++) {
		if (n % i == 0) {
			int temp = n / i;
			if (isPrime(i) && isPrime(temp)) {
				cout << max(i, temp) << endl;
				break;
			}
		}
	}

	return 0;
}

哥德巴赫猜想

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

bool isPrime(int x)
{
	if (x < 2)return false;
	if (x == 2)return true;
	for (int i = 2; i <= sqrt(x); i++) {
		if (x % i == 0)return false;
	}
	return true;
}

int main()
{
	int n;
	cin >> n;

	for (int i = 4; i <= n; i+=2) {
		for (int j = 2; j < n; j++) {
			int temp = i - j;
			if (isPrime(j) && isPrime(temp)) {
				cout << i << "=" << j << "+" << temp << endl;
				break;
			}
		}
	}


	return 0;
}

[USACO1.5] 回文质数 Prime Palindromes

如果用字符串来判断是否为回文数字，会超时：

bool isPalindrome(string x)
{
	for (int i = 0, j = x.length() - 1; i < j; i++, j--) {
		if (x[i] != x[j])return false;
	}
	return true;
}

解决方法一：把数字反转，而不是用字符串

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

bool isPalindrome(int x)
{
	int n = x;
	int reverse = 0;
	while (n > 0) {
		reverse = reverse * 10 + n % 10;
		n /= 10;
	}
	if (reverse == x)return true;
	return false;
}

bool isPrime(int x)
{
	for (int i = 2; i <= sqrt(x); i++) {
		if (x % i == 0)return false;
	}
	return true;
}


int main()
{
	int low, high;
	cin >> low >> high;
	for (int i = low; i <= high; i++) {
		if (isPalindrome(i) && isPrime(i)) {
			cout << i << endl;
		}
	}


	return 0;
}

解决方法二：判断素数的方法换成埃拉托斯特尼筛法

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
using namespace std;

bool isPalindrome(string x)
{
	for (int i = 0, j = x.length() - 1; i < j; i++, j--) {
		if (x[i] != x[j])return false;
	}
	return true;
}

// 普通埃拉托斯特尼筛法，找出 2 到 sqrt(b) 范围内的所有质数
void sieve(int limit, vector<int>& primes) {
    vector<bool> isPrime(limit + 1, true);
    isPrime[0] = isPrime[1] = false;  // 0和1不是质数

    for (int i = 2; i <= limit; i++) {
        if (isPrime[i]) {
            primes.push_back(i);
            for (int j = i * i; j <= limit; j += i) {
                isPrime[j] = false;
            }
        }
    }
}

// 分段筛法，找出区间 [low, high] 内的质数
void segmentedSieve(int low, int high) {
    // 计算最大值 high 的平方根
    int limit = sqrt(high) + 1;

    // 获取小于等于 sqrt(high) 的所有质数
    vector<int> primes;
    sieve(limit, primes);

    // 创建一个区间 [low, high] 的布尔数组，初始化为 true（表示是质数）
    vector<bool> isPrime(high - low + 1, true);

    // 对于每一个小于等于 sqrt(high) 的质数，标记出在 [low, high] 范围内的倍数
    for (int p : primes) {
        // 找到 [low, high] 范围内的 p 的第一个倍数
        int start = max(p * p, (low + p - 1) / p * p); // p^2 或者从低范围开始的倍数

        // 标记 [low, high] 范围内的 p 的倍数为非质数
        for (int j = start; j <= high; j += p) {
            isPrime[j - low] = false;
        }
    }

    // 输出 [low, high] 范围内的回文质数
    for (int i = 0; i <= high - low; i++) {
        if (isPrime[i] && (low + i) >= 2 && isPalindrome(to_string(low + i))) {
            cout << low + i << endl;
        }
    }
}

int main()
{
	int low, high;
	cin >> low >> high;

	
    // 使用分段筛法查找 [a, b] 范围内的回文质数
    segmentedSieve(low, high);


	return 0;
}

集合求和

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
using namespace std;

int a[31];

int main()
{
	int i = 0;
	long long sum = 0;
	while (cin >> a[i++]);
	for (int j = 0; j < i; j++) {
		sum += a[j];
	}
	sum *= pow(2, i - 2);
	cout << sum << endl;

	return 0;
}

注意：

这里需要用到一个推导，，推导过程大家可以自己去网上查找

【深基7.习8】猴子吃桃

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
using namespace std;

int main()
{
	int day;
	cin >> day;
	int n = 1;
	for (int i = 1; i < day; i++) {
		n = (n + 1) * 2;
	}
	cout << n << endl;

	return 0;
}

【深基7.习9】培训

#include <iostream>
#include <cstring>
#include <string>
#include <cmath>
#include <vector>
using namespace std;

struct info
{
	string name;
	int age;
	int score;
}a[10];

int main()
{
	int n;
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i].name >> a[i].age >> a[i].score;
		if (1.2 * a[i].score <= 600)a[i].score *= 1.2;
		else a[i].score = 600;
	}
	for (int i = 0; i < n; i++) {
		cout << a[i].name << " " << a[i].age + 1 << " " << a[i].score << endl;
	}


	return 0;
}