【算法通关指南：C++编程篇（五） 】---- 条件判断与循环练习合集（下）

#include <iostream>
using namespace std;
typedef long long LL;
LL m,k;

int main()
{
	cin >> m >> k;

	int ret = 0;
	while (m)
	{
		if (m % 10 == 3)
			ret++;
		m /= 10;
	}

	if (ret == k)
		cout << "YES" << endl;
	else
		cout << "NO" << endl;
	return 0;
}

#include <iostream>
using namespace std;

int main()
{
	long long n;
	scanf("%lld", &n);

	while (n != 1)
	{

		if (n % 2 != 0)
		{
			printf("%lld*3+1=%lld\n", n,n * 3 + 1);
			n = n * 3 + 1;
		}
		else if(n % 2 == 0)
		{
			printf("%lld/2=%lld\n", n, n / 2);
			n /= 2;
		}
	}
	printf("End\n");
	return 0;
}

#include <iostream>
using namespace std;

int main()
{
	int sum = 0;

	for (int i = 1; i <= 100; i++)
	{
		if (i % 3 == 0)
			sum += i;
	}

	cout << sum << endl;
	return 0;
}

#include <iostream>
using namespace std;

int main()
{
	int sum = 0;

	for (int i = 3; i <= 100; i += 3)
			sum += i;

	cout << sum << endl;
	return 0;
}

#include <iostream>
using namespace std;
int main()
{
	int n;
	cin >> n;

	int sum = 0;
	for (int i = 1; i <= n; i++)
	{
		int x;
		cin >> x;
		sum += x;
	}
	double ret = sum * 1.0 / n;

	printf("%.2lf", ret);
	return 0;
}

#include<iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    int a,b,c;
    int sum_a = 0;
    int sum_b = 0;
    int sum_c = 0;
    int sum = 0;
    for(int i = 0;i < n;i++)
    {
        cin >> a >> b >> c;
        sum_a += a;
        sum_b += b;
        sum_c += c;
    }
    sum = sum_a + sum_b + sum_c;
    cout << sum_a << " " << sum_b << " " << sum_c << " " << sum;
    return 0;
}

#include <iostream>
using namespace std;
int n;
int main()
{
	cin >> n;

	int a,b;
	cin >> a >> b;
	double x = (b * 1.0 / a) * 100;

	for (int i = 1; i < n; i++)
	{
		cin >> a >> b;
		double y = (b * 1.0 / a) * 100;
		if (y - x > 5)
			cout << "better" << endl;
		else if (x - y > 5)
			cout << "worse" << endl;
		else
			cout << "same" << endl;
	}
	return 0;
}

#include <iostream>
using namespace std;

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

	double s  = 0;
	int fact = 1;
	for (int i = 1; i <= n; i++)
	{
		s += 1.0 / i * fact;
		fact *= -1;
	}

	printf("%.4lf", s);
	return 0;
}

#include <iostream>
using namespace std;

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

	int p = 1;
	int q = 2;
	double sum = 0;
	for (int i = 1; i <= n; i++)
	{
		sum += (q * 1.0/ p);
		q = q + p;
		p = q - p;
	}
	
	printf("%.4lf", sum);
	return 0;
}

#include <iostream>
using namespace std;

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

	int ans = 0;
	do
	{
		ans++;
		n /= 10;

	} while (n);
	cout << ans << endl;
	return 0;
}