this is my code for this problem and it worked well
#include <iostream>
#include <vector>
#include <string>
using namespace std;
void changevalue(pair<int, int> p, string s, int j);
int main() {
int T, N, countE = 0, countS = 0;
string lydia;
string out = "";
vector<string> output(2);
cin >> T;
output.resize(T);
for (int i = 0; i < T; i++) {
cin >> N;
cin >> lydia;
pair<int, int> lyd;
pair<int, int> me;
lyd = { 0, 0 };
me = { 0,0 };
for (int j = 0; j < 2 * N - 2; j++) {
if ((me.first == me.second) && (me.first == N - 1))
break;
if (lyd == me) {
out += 'E' + 'S' - lydia[j];
}
else {
if (out[j - 1] == 'E')
out += 'S';
else out += 'E';
}
changevalue(lyd, lydia, j);
changevalue(me, out, j);
}
output.at(i) = out;
out = "";
}
for (int k = 0; k < T; k++) {
cout << "Case #" << k + 1 << ": " << output.at(k) << endl;
}
system("pause");
return 0;
}
void changevalue(pair<int, int> p, string s, int j) {
if (s[j] == 'E') p.second++;
if (s[j] == 'S') p.first++;
}
Le lun. 8 avr. 2019 à 22:41, Gokul Sreeram <gokulfalconsree...@gmail.com> a
écrit :
> On Monday, 8 April 2019 23:09:08 UTC+5:30, Aabhas wrote:
> > This is a good approach.
> > It just requires to check the move every step and make sure limits are
> not crossed.
> >
> >
> >
> > On Mon, Apr 8, 2019 at 9:03 AM Luke Pebody <lu...@pebody.org> wrote:
> >
> > Mine was to walk down the diagonal making sure that each time you
> stopped off the diagonal you were the opposite side of the diagonal from
> Angela or whatever the other walker was called.
> >
> >
> > On Sun, 7 Apr 2019, 10:36 pm Ronan Burke, <burk...@gmail.com> wrote:
> > My solution for this one was to just do the opposite of the route the
> other person took. If you did something other than, I imagine things could
> get quite messy and it would be easy to make a mistake.
> >
> >
> >
> > Their route: "SSEESE" then I would do "EESSES"
> >
> >
> >
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> >
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> >
> >
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> >
> > Cheers!
> > Aabhas
>
>
> Hi,
>
> In my code. There is no possibilities of limits gets exceeding.. If
> possible please let me know if there is any failing cases
>
> Regards,
> Gokul Sreeram
>
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