Codeforces412EEmailAddresses(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
#include
#include
#include
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 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 if (s[j] == ‘@‘ || s[j] == ‘_‘) {
i = j-1;
break;
}
if (s[j] == ‘.‘) {
if (j - i == 1) break;
int k;
for (k = j+1; k if (!isLetter(s[k]))
break;
y++;
}
i = k-1;
break;
}
}
ans += x * y;
}
}
cout < return 0;
}

Codeforces 412E E-mail Addresses(数论),布布扣,bubuko.com