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$A \sim J$には，各々のアルファベットに$0 \sim 9$が対応する．全ての和の最大値を答える．まずは各アルファベットを1にして考えて，アルファベットごとの和を出す．この和が大きいものから数字を大きい順に当てはめていく．もし0で始まってしまう場合があったらそこはswapする．
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#include <iostream>
#include <sstream>
#include <vector>
#include <string>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#define REP(i,k,n) for(int i=k;i<int(n);i++)
#define rep(i,n) for(int i=0;i<int(n);i++)
#define each(it,v) for(__typeof((v).begin()) it=(v).begin();it!=(v).end();it++)
#define INF 1<<30
#define mp make_pair
using namespace std ;
typedef long long ll ;
typedef pair < ll , ll > P ;
class EncodedSum {
public :
long long maximumSum ( vector < string > numbers ) {
vector < vector < int > > s ( numbers . size ());
rep ( i , numbers . size ()) {
s [ i ]. resize ( numbers [ i ]. size ());
rep ( j , numbers [ i ]. size ()) {
s [ i ][ j ] = int ( numbers [ i ][ j ] - 'A' );
}
}
bool flag [ 10 ];
memset ( flag , 0 , sizeof ( flag ));
vector < ll > x ( 10 );
rep ( i , s . size ()) {
ll t = 1 ;
flag [ s [ i ][ 0 ] ] = true ;
for ( int j = s [ i ]. size () - 1 ; j >= 0 ; j -- ) {
x [ s [ i ][ j ] ] += t ;
t *= 10 ;
}
}
vector < P > v ;
vector < int > a ( 10 );
rep ( i , 10 ) {
v . push_back ( P ( x [ i ], i ));
a [ i ] = 9 - i ;
}
sort ( v . begin (), v . end (), greater < P > ());
ll ans = 0 ;
for ( int i = v . size () - 1 ; i >= 0 ; i -- ) {
if ( flag [ v [ i ]. second ] && a [ i ] == 0 ) {
swap ( a [ i ], a [ i - 1 ]);
}
ans += v [ i ]. first * a [ i ];
}
return ans ;
}
};