이 제출은 이전 버전의 oj.uz에서 채점하였습니다. 현재는 제출 당시와는 다른 서버에서 채점을 하기 때문에, 다시 제출하면 결과가 달라질 수도 있습니다.
#include "dna.h"
#include <iostream>
using namespace std;
#define nmax 100001
int n;
int AtoT[nmax], AtoC[nmax], TtoA[nmax], TtoC[nmax], CtoA[nmax], CtoT[nmax];
int valAtoT, valAtoC, valTtoA, valTtoC, valCtoA, valCtoT;
string A, B;
void init( string a, string b ) {
int i;
n = a.size();
A = a;
B = b;
a = '0' + a;
b = '0' + b;
for( i = 1; i <= n; i++ ) {
AtoT[i] = AtoT[i - 1];
AtoC[i] = AtoC[i - 1];
TtoA[i] = TtoA[i - 1];
TtoC[i] = TtoC[i - 1];
CtoA[i] = CtoA[i - 1];
CtoT[i] = CtoT[i - 1];
if( a[i] == b[i] )
continue;
if( a[i] == 'A' ) {
if( b[i] == 'T' )
AtoT[i]++;
else
AtoC[i]++;
} else if( a[i] == 'C' ) {
if( b[i] == 'T' )
CtoT[i]++;
else
CtoA[i]++;
} else {
if( b[i] == 'A' )
TtoA[i]++;
else
TtoC[i]++;
}
}
}
int get_distance( int x, int y ) {
int minn, rez = 0;
x++;
y++;
valAtoC = AtoC[y] - AtoC[x - 1];
valAtoT = AtoT[y] - AtoT[x - 1];
valTtoC = TtoC[y] - TtoC[x - 1];
valTtoA = TtoA[y] - TtoA[x - 1];
valCtoA = CtoA[y] - CtoA[x - 1];
valCtoT = CtoT[y] - CtoT[x - 1];
minn = min( valAtoC, valCtoA );
valAtoC -= minn;
valCtoA -= minn;
rez += minn;
minn = min( valAtoT, valTtoA );
valAtoT -= minn;
valTtoA -= minn;
rez += minn;
minn = min( valCtoT, valTtoC );
valCtoT -= minn;
valTtoC -= minn;
rez += minn;
//cout << "rez " << rez << "\n";
if( valCtoA == valAtoT && valAtoT == valTtoC && valCtoT == valTtoA && valTtoA == valAtoC )
return rez + 2 * valCtoA + 2 * valAtoC;
return -1;
}
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