Codeforces 412E E-mail Addresses(数论)

题目链接：Codeforces 412E E-mail Addresses

题目大意：问说给出字符串中有几个正确的邮箱，

1.‘@’前非空的字符串，由字符，数字和下划线组成，只能由字符开头

2.’@‘和’.‘中间的字符串非空，只能由字符，数字和下划线组成

3.‘.’后非空字符串，只能有字符组成

解题思路：计算每个@前面字符的个数，以及.后面字符的个数，相乘即为当前这对@和.所满足的个数，不过要注意@和.之间的字符串非空。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>

using namespace std;
typedef long long ll;
const int N = 1e6+5;

char s[N];

bool isLetter(char c) {
	if (c >= ‘a‘ && c <= ‘z‘) return true;
	if (c >= ‘A‘ && c <= ‘Z‘) return true;
	return false;
}

int main () {
	scanf("%s", s);

	int n = strlen(s);
	ll ans = 0;
	for (int i = 0; i < n; i++) {
		if (s[i] == ‘@‘) {
			ll x = 0, y = 0;
			for (int j = i-1; j >= 0; j--) {
				if (s[j] == ‘@‘ || s[j] == ‘.‘)
					break;
				if (isLetter(s[j]))
					x++;;
			}

			for (int j = i+1; j < n; j++) {
				if (s[j] == ‘@‘ || s[j] == ‘_‘) {
					i = j-1;
					break;
				}

				if (s[j] == ‘.‘) {
					if (j - i == 1) break;
					int k;
					for (k = j+1; k < n; k++) {
						if (!isLetter(s[k]))
							break;
						y++;
					}
					i = k-1;
					break;
				}
			}
			ans += x * y;
		}
	}
	cout << ans << endl;;
	return 0;
}

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Codeforces 412E E-mail Addresses(数论)

原文：http://blog.csdn.net/keshuai19940722/article/details/24179639