c++字符串和日期基础

一，字母三角形

#include<string>
#include<iostream>
using namespace std;
int main()
{
	int n = 0;
	cin >> n;
	for (int i = 1; i <= n; i++)//i代表行数
	{
		string space=string(n - i, ' ');//前半部分空格
		string ch = string(2 * i - 1, 'A' + i - 1);
		cout << space + ch << endl;
	}
	return 0;//cpp文件一定要return 0
}

2，升级三角形

#include<string>
#include<iostream>
using namespace std;
int main()
{
	char c;
	cin >> c;
	if (c >= 'A' && c <= 'Z')
	{
		for (int i = 1; i <= c - 'A' + 1; i++)
		{
			for (int j = 1; j <= c - 'A' + 1 - i; j++)
			{
				cout << " ";
			}
			for (int j = 1; j <= i; j++)
				cout << (char)('A' + j-1);//输出字符要在前方括起来
			for(int j=i-1;j>0;j--)
				cout << (char)('A' + j-1);
			cout << endl;

	}

	}
	else if (c >= '0' && c <= '9')
	{
		for (int i = 1; i <= c - '1' + 1; i++)
		{
			for (int j = 1; j <= c - '1' + 1 - i; j++)
			{
				cout << " ";
			}
			for (int j = 1; j <= i; j++)
				cout << (char)('1' + j - 1);//输出字符要在前方括起来
			for (int j = i - 1; j > 0; j--)
				cout << (char)('1' + j - 1);
			cout << endl;

		}
	}
	return 0;//cpp文件一定要return 0
}

3，寻找字符串

#include<cstdio>
#include<cstring>
char s1[1005], s2[1005];
int main()
{
	fgets(s1, 1004, stdin);
	fgets(s2, 1004, stdin);
	int ans = 0; 
	int l1 = strlen(s1)-1; int l2 = strlen(s2)-1;

	for ( int i = 0; l2 + i-1 < l1; i++)
	{
		bool mach = true;
		for (int j = 0; j< l2; j++)
		{
			if (s1[i + j] != s2[j])
			{
				mach = false;
				break;//结束循环从下一个字符开始寻找
			}
		}
		if (mach) {
			ans++;
		}
	}
	printf("%d", ans);
	return 0;
}

4，日期计算

#include<iostream>
#include<string>
using namespace std;
int whatday(int y, int m, int d)
{//分析 2019 11 29日   前面有完整的2018个年份，1-11月加28天
	//返回正确的星期，用0-6表示1-7
	int nas= 0;
	for (int i = 1; i < y; i++)//当年还没有过完，要减1
	{
		if ((i % 4 == 0 && i % 100 != 0 )|| i % 400 == 0)//闰年判断条件
		{
			nas += 366%7;
			nas%=7;
		}
		else {
			nas += 365 % 7;
			nas %= 7;
		}
	}
	for (int i = 1; i < m; i++)
	{
		if (i == 1 || i == 3 || i == 5 || i == 7 || i == 8 || i == 10 || i == 12)
		{
			nas += 31 % 7;
			nas %= 7;
		}
		else if (i == 4 || i == 6 || i == 9 || i == 11 )//一定要用else if构成并列关系，否则if下面又有else if，只有下半部分发生并列
		{
			nas += 30 % 7;
			nas %= 7;
		}
		else if ((y % 4 == 0 && y % 100 != 0 )|| y % 400 == 0)
		{
			nas += 29%7;
			nas %= 7;
		}
		else
		{
			nas += 28 % 7;
			nas = nas % 7;
		}
	}
	
		nas += d - 1;//关于为什么是-1，不减1则余数顺序为1234567 对应1，2，3，4，5，6，0，减一后变为0，1，2，3，4，5，6，符合下标
		//天数不需要循环了
	return nas%7;

}
string weekday[7] = { "Monday","Tusday","Wednesday","Thursday","Friday","Saturday","Sunday"};
int main()
{
	int y, m, d;
	cin >> y >> m >> d;
	cout << weekday[whatday(y, m, d)] << endl;
	return 0;
}

或者：

int whatday(int y,int m,int d){
if(m<=2){
m+=12;
y--;}
return (d+2*m+3*(m+1)/5+y+y/4-y/100+y/400)%7;}