Codeforces1045I

Posted 2021-01-09 rrrr-wys

Codeforces1045I

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做法：如果至多有一种字母出现奇数次，则合法，因此把整个串用一个26位2进制数表示，对于每一个数，枚举可能的对应串的状态即可。

#include <bits/stdc++.h>
typedef long long ll;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while( ! isdigit(c) ) { if(c == ‘-‘) f = -1; c = getchar(); }
    while( isdigit(c) ) { x = x*10 + c - ‘0‘; c = getchar(); }
    return x * f;
}
inline void write( ll x ) {
    if( x >= 10 ) write( x / 10 );
    putchar( x % 10 + ‘0‘ );
}
using namespace std;
int n, w;
char c;
ll ans = 0;
map< ll , int > M;
int main() {
    n = read();
    for ( int i = 1 ; i <= n ; ++ i ) {
        w = 0;
        c = getchar();
        while( c != ‘
‘ ) { w ^= ( 1 << ( c - ‘a‘ ) ); c = getchar(); }
        ans += M[ w ];
        for ( int j = 0 ; j < 26 ; ++ j ) 
            if( M.find( w ^ ( 1 << j ) ) != M.end() ) 
                ans += M[ w ^ ( 1 << j ) ];
        ++ M[ w ];
    }
    write( ans ); putchar( ‘
‘ );
    return 0;
}