#include<bits/stdc++.h>

using namespace std;

const double pp = acos(-1);     // 等于 pi, 对精度要求高的题目有用

int main()

{

	int a, b;

	cin >> a >> b;

	int c = (b + a) * (b + a);

	printf("%0.10lf", pp* (double)c);

	return 0;

}

#include <bits/stdc++.h>

#include<iostream>

using namespace std;

int main()

{

    string s;

    cin>>s;

    int ans=0;

    int a=0,b=0;

    for(int i=0;i<s.size();i++) {

        if(s[i]=='a') a++;

        else b++;

        if(a%10==0) ans++;

        if(a<10) ans++;

        if(b%10==0) ans++;

        if(b<10) ans++;

    }

    printf("%d",ans);

}

#include <iostream>

#include <stdio.h>

using namespace std;

int a[13] = { 0,31,28,31,30,31,30,31,31,30,31,30,31 };

int y[2] = { 365,366 };

bool judge(int i) {

	if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0) {

		return true;

	}

	return false;

}

int main()

{

	int yy, mm, dd;

	cin >> yy >> mm >> dd;

	int sum = 0;

	for (int i = 2022; i < yy; i++) {

			if (judge(i)) sum += y[1];

			else sum += y[0];

	}

	for (int i = 1; i < mm; i++) {

		if (judge(yy)) a[2] = 29;

		sum += a[i];

	}

	sum += dd;

	cout << sum;

	return 0;

}

#include <iostream>

#include <stdio.h>

using namespace std;

int main()

{

	int n;

	cin >> n;

	int res=0;

	int sum = 0;

	for (int i = 1; i <= n; i++) {

		int num = i;

		while (num) {

			res = num % 10;

			if (res == 2 || res == 1 || res == 9 || res == 0) {

				sum += i;

				break;

			}

			num /= 10;

		}

	}

	cout << sum;

	return 0;

}

20110101

20111231

1

不能枚举日期然后进行判断，这样跟写一个日历似的，太复杂

枚举所有年份，然后把它翻过来扩充成一个“日期”，然后看这个得到的“日期”是不是一个正常的日期格式，然后在看看是不是符合输入给的范围

 check函数来判断这个日期是不是一个正常的日期格式

一个数除几个0，就是忽略掉这个数后面的几位

一个数mod几个0，就是拿到这个数后面的几位

#include <cstring>

#include <iostream>

using namespace std;

int days[13] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};

bool check(int date)

{

    int year = date / 10000;

    int mouth = (date / 100) % 100;

    int day = date % 100;

    if(mouth < 0 || mouth > 12) return false;

    if(day == 0 || mouth != 2 && day > days[mouth]) return false;

    if(mouth == 2)

    {

        int leap = year % 100 && year % 4 == 0 || year % 400 == 0;

        if(day > days[mouth] + leap)   return false;

    }

    return true;

}

int main()

{

    int date1, date2;

    cin >> date1 >> date2;

    int res = 0;

    for(int i = 1000; i < 10000; i ++)

    {

        int date = i, x = i;

        for(int j = 0; j < 4; j ++ )    date = date * 10 + x % 10, x /= 10;

        if(date1 <= date && date <= date2 && check(date))   res ++;

    }

    cout << res << endl;

    return 0;

}