{
dp[i][j] = min(dp[i][j], dp[i-1][k] + (!isEven[(~j) & (c[i] ^ k)] + num[j & (c[i] ^ k)] + ((num[j ^ (j & (c[i] ^ k))] - num[c[i] & k]) << 1)) * (1 << (i-1)));
}
}
}
}
}
}
int ans = INF;
for(int i=0;i<(1<
{
ans = min(ans, dp[m][i]);
}
}
printf("%d\n", ans);
}
}
return 0;
}
作者：CyberZHG